[V.5] Solution Book for
by Prof Sang-Gu LEE with SKKU Students
http://matrix.skku.ac.kr/LA-Lab/Solution/
http://matrix.skku.ac.kr/2015-Album/Big-Book-LinearAlgebra-Eng-2015.pdf
(Based on the book written by Sang-Gu Lee with Jae Hwa Lee, Kyung-Won Kim)
http://matrix.skku.ac.kr/2015-Album/BigBook-LinearAlgebra-2015.pdf
http://matrix.skku.ac.kr/CLA-Exams-Sol.pdf,
http://matrix.skku.ac.kr/2015-album/2015-LA-S-Exam-All-Sol.pdf
http://matrix.skku.ac.kr/2015-album/2015-LA-F-Midterm-Final-Solution-F.pdf
Made by SGLee http://matrix.skku.ac.kr/sglee/
2015.9.8. Solved by 박시현 2015.9.8. Revised by 정용석 2015.9.12.
Revised by 변희성 2015.11.10. Finalized by 김준현 Final OK by SGLee
For points , find the vector .
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2505/ by 변희성
INPUT
P1=vector([5,-2,1])
P2=vector([2,4,2])
vec=P2-P1
print vec
OUTPUT
(-3, 6, 1) : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.8. Solved by 박시현 2015.9.8. Revised by 정용석 2015.9.12.
Revised by 변희성 2015.11.10. Finalized by 김준현 Final OK by SGLee
What is the initial point of the vector with terminal point ?
Sol) Initial point
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2506/ by 변희성
INPUT
x=vector([1,1,3])
B=vector([-1,-1,2])
point=B-x
print point
OUTPUT
(-2, -2, -1) : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.10. Solved by 김민기 2015.9.30. Finallized by 김민기 2015.9.30. Refinallized by 김민기 Final OK by SGLee
(NEW )
What is the initial point of the vector 3 2 1 with terminal point
2 3 8 ?
Sol)
■
Checked by Sage) http://math3.skku.ac.kr/home/pub/23 by 김민기
INPUT
x = vector([3, 2, 1])
B = vector([2, 3, 8])
print "vector xB ", B-x
OUTPUT
vector xB (-1, 1, 7) : OK ■
2015.9.8. Solved by 박시현 2015.9.8. Revised by 정용석 2015.9.12.
Revised by 변희성 2015.11.10. Finalized by 김준현 Final OK by SGLee
For vectors , , and
, compute the following:
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2507/ by 변희성
INPUT
u=vector([-3,1,2,4,4])
v=vector([4,0,-8,1,2])
w=vector([6,-1,-4,3,-5])
ans=(2*u-7*w)-(8*v+u)
print ans
OUTPUT
(-77, 8, 94, -25, 23) : OK ■
2015.9.8. Solved by 전덕우 2015.9.9. Revised by 강민성 2015.9.9.
Finalized by 김원경 2015.9.15. Refinalized by 김원경 Final OK by SGLee
Using the same vectors and from above, find the vector that
satisfies the following:
Sol) Let
⇒
⇒
Checked by Sage) http://math1.skku.ac.kr/home/pub/2447 by 김원경
INPUT
u=vector([-3,1,2,4,4])
v=vector([4,0,-8,1,2])
w=vector([6,-1,-4,3,-5])
x=(2*u-v-w)/6
print x
OUTPUT
(-8/3, 1/2, 8/3, 2/3, 11/6) : OK ■
2015. 9. 9. Solved by 배성준 2015. 9. 11 Revised by 정양헌 2015. 9. 12. Finalized by 서승완
2015. 9. 16. ReFinalized by 배성준 2015. 10. 12. Refinalized by 박시현 Final OK by SGLee
(New)
For vectors , , and
. Find the vector that satisfies the following:
Sol) =>
=>
=>
=>
∴ ■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2427 by 배성준
INPUT
u=vector([-3,1,2,4,4])
v=vector([4,0,-8,1,2])
w=vector([6,-1,-4,3,-5])
x=(2*u-3*v-w)/8
print"x=",x
OUTPUT
x= (-3, 3/8, 4, 1/4, 7/8) : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.9. Solved by 김성찬 2015.9.9. Finalized by 김성찬 2015.9.13. Refinalized by 정다산 2015.9.23. Refinalized by 김성찬, 임상훈
2015.9.29. Final OK by 이경승 Final OK by SGLe
For vectors calculate , where is
the angle between x and y.
Sol)
.
. ■
Note : ∴
Checked by Sage) http://math1.skku.ac.kr/home/pub/2508/ by 임상훈
INPUT
x=vector([-1,-2,3])
y=vector([3,-2,-1])
print "cos(theta)= ", x.inner_product(y)/(x.norm()*y.norm())
OUTPUT
cos(theta)= -1/7 : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.11.10. Solved by 김준현 2015.11.10. Finalized by 정양헌 Final OK by SGLee
Find the distance between the two points and .
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2509 by 정양헌
INPUT
P=vector([-1,2,1])
Q=vector([-3,-4,5])
print "P-Q =",(P-Q).norm()
OUTPUT
P-Q = 2*sqrt(14) : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.8. Solved by 정범진 2015.9.8. Revised by SGLee 2015.9.8. Revised by 김원경
2015.9.9. Revised by 강민성 2015.9.11. Finalized by 김원경 2015.9.16. Refinalized by 정범진
2015.11.10. Refinalized by 정양헌 Final OK by SGLee
For vectors , , find the real number
that satisfy .
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2510/ by 정양현
INPUT
a=var('a')
x=vector([a,2,-1,a])
y=vector([-a,-1,3,6])
xy=[x[i]*y[i] for i in range(4)]
solve (sum(xy), a)
OUTPUT
[a == 1, a == 5] : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.10. Solved by 김민기 2015.9.30. Finallized by 김민기 2015.10.12. Refinalized by 박시현
Final OK by SGLee
(New)
For vectors x= (a, 2, -1, a) , y = (a, 2, 3, 2), find the real number a such that x . y = 0.
Sol)
■
Checked by Sage) http://math3.skku.ac.kr/home/pub/41 by 박시현
INPUT
a=var('a')
x=vector([a,2,-1,a])
y=vector([a,2,3,2])
xy=[x[i]*y[i] for i in range(4)]
print xy
solve ([a^2+2*a+1],a)
OUTPUT
[a^2, 4, -3, 2*a]
[a == -1] : OK ■
2015.9.12. Solved by 호재원 2015.9.12 Revised by 권순호 2015.9.12 Finalized by 김동욱
10.08 Refinalized by 김동욱 Final OK by SGLee
Find a vector equation of the line between the two points .
Sol) A equation of line passes through the point
either parallel to a vector
= ( )
∴ , () ■
Check by Sage) http://math1.skku.ac.kr/home/pub/2488 by 주민규
INPUT
p=vector([-5,1,3])
q=vector([2,-3,4])
t=real
x=q-p
print x
OUTPUT
(7, -4, 1) : OK ■
NOTE)
We can use this vector to find the equation of line by adding a vector to equation multiplied by a real number .
2015.9.10. Solved by 김민기 2015.9.30. Finallized by 김민기 2015.9.30. Refinallized by 김민기 Final OK by SGLee
2015.11.19. Refinalized by 주민규 Final OK by SGLee
(New)
Find a vector equation of the line between the two points
and .
Sol)
The vector equation pass through two points and :
or
∴ ■
Check by Sage) http://math3.skku.ac.kr/home/pub/126 by 주민규
INPUT
var('x,y,z')
var('t')
P = vector([5,0,1])
Q = vector([0,2,6])
R= Q-P
print "vector equation= ", P+R*t
OUTPUT
vector equation= (-5*t + 5, 2*t, 5*t + 1) : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.9. Solved by 김성찬 2015.9.11. Finalized by 김성찬 2015.9.14. Refinalized by 주민규
Final OK by SGLee
Find a normal vector perpendicular to the plane .
Sol) Normal vector : Length of vector is 1.
.
(perpendicular vector to the plane)
.
■
Finalized by 정양헌 Final OK by SGLee
[Projection] For and , find the scalar projection and vector projection of onto .
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2511/ by 정양현
INPUT
a=vector([2, -1, 3])
b=vector([4, -1, 2])
ab=a.inner_product(b)
aa=a.inner_product(a)
p=ab/aa*a;w=b-p
print "p=", p
print "w=", w
OUTPUT
p= (15/7, -15/14, 45/14)
w= (13/7, 1/14, -17/14) :OK ■
You can practice this code in http://sage.skku.edu/ .
2015.11.19. Solved by 주민규 2015.11.23. Revised by 정양헌 Final OK by SGLee
[Discussion] Vectors with the same magnitude and direction are considered to be equivalent.
However, in a vector space, discuss the relationship between vectors with the same slope but expressed with different equations.
Ans)
Equivalence
When each components of two vectors are same, then we say
that and are equivalent (or equal) and we write .
[the relationship between vectors with the same slope but expressed with different equations.]
Conclusively, if the direction of two lines and the each slop are same, then it means one line can be obtained by moving (parallel) the other line and change the length. After that they can be considered as one same vector(line). ■
Note: We can define a line in two dimensional space as (, is a position vector for any given point on the line, is a position vector to any known point on the line and is a direction vector parallel to the line).
If we choose same but different then we get the two equations of line in a vector space with the same slope but expressed with different equations.
For example we can consider the line with and then we get , and we can consider the other line with same but different then we can get . From this example we can know that there are some vectors with the same slope but expressed with different equations. ∎
(PS: 결론적으로, 두 선분이 기울기와 방향이 같다면, 하나의 선분을 평행이동하여 길이만 늘려주면 같은 벡터(선분)가 되는 관계가 있다. )
2015.9.8. Solved by 이승열 2015.9.8. Finalized by 김원경 2015.9.14.
Refinalized by 주민규 Final OK by SGLee
Sol)
Check : and
=> : orthonormal vectors
[Find which is orthonormal to both and .]
=>
=> , .
We know that for some .
In addition, must be a normal vector which means that =>
Answer: .
2015.9.19. Solved by 서승완, 2015.9.19. Revised by 이반석, 김성찬, 2015.9.29. Final OK by 이경승 Final OK by SGLee
Answer the questions for the following linear system.
(1) Find the coefficient matrix.
(2)Express the linear system in the form .
(3) Find it’s augmented matrix.
Solution.
(1) The coefficient matrix of the following linear system is .
(2) Let , , . So the following linear system can be expressed by .
(3) It’s augmented matrix is ■
Sage http://math3.skku.ac.kr/home/pub/15
A=matrix(3, 3, [1,3,-1,2,5,1,1,1,1])
b=vector([1,5,3])
print A.augment(b,subdivide=True)
[ 1 3 -1| 1]
[ 2 5 1| 5]
[ 1 1 1| 3] Okay
You can practice this code in http://sage.skku.edu/ .
2015.9.12. Solved by 서승완 2015.9.12. Revised by 남기현 2015.9.30. Finalized by 남기현
Find a linear system with its augmented matrix.
(Put the unknowns as .)
Solution.
Let be the coefficient matrix, the unknown, and b the constant.
then = , =
Hence we have
<=> =
∴ Therefore, linear system is .
Note:
This problem does not need to use Sage. ■
2015.10.29. Solved by 주영은, 2015.10.29. Revised by 서승완, 변희성, 주민규, ReFinalized and Final Ok by SGLee
Find the number of leading variables and free variables in the solution set of the following system.
SOL> Let , = .
= RREF
Leading variables are , , , and free variables are , . ■
Double checked by Sage > http://math3.skku.ac.kr/home/pub/69 by 주영은
A = matrix(3,6,[1,4,5,-9,7,1,0,2,4,-6,-6,2,0,0,0,-5,0,3])
print A.rref()
[ 1 0 -3 0 19 -6/5]
[ 0 1 2 0 -3 -4/5]
[ 0 0 0 1 0 –3/5] OK. ■
You can practice this code in http://sage.skku.edu/ .
2015.9.19. Solved by 서승완, 2015.09.23. Revised by 김성찬
Which matrices are in REF or RREF? If one is not in RREF, transform it to a form in RREF.
, .
Sol) Let the matrices , .
At the matix row’s first number(except 0) is not zero. So it is not REF and also RREF too. At the matix B row’s first number(except 0)is in the same column with row. So it is not REF and RREF.
For the matrix ,
,
For the matrix B,
.
Answer : There are nor REF and RREF.
For each matrix RREF is , . ■ Sage http://math3.skku.ac.kr/home/pub/14
A=matrix(3,4,[1,2,-1,-2,0,2,-2,-3,0,0,0,2])
B=matrix(3,5,[0,1,0,0,5,0,0,1,0,4,0,1,0,-2,3])
print "RREF(A)="
print A.rref()
print "RREF(B)="
print B.rref()
RREF(A)= RREF(B)=
[ 1 0 1 0] [0 1 0 0 5]
[ 0 1 –1 0] [0 0 1 0 4] Okay
You can practice this code in http://sage.skku.edu/ .
2015.9.12. Solved by 김준현 2015.9.12. Revised by 정진규, 2015.9.13. Revised by 남기현, 2015.9.14. Finalized by 김준혁, 2015.10.11. Refinalized by 정진규, Final OK by SGLee
Solve the system using Gaussian elimination.
Sol)
The augmented matrix of the system can be expressed as below.
Find
However, at , the last equation in the system is and it is impossible. Therefore, there’s no answer for the system. ■
2015.9.11. Solved by 전덕우 2015.9.11. Revised by 정양헌 2015.9.12. Finalized by 김원경 2015.9.14. Refinalized by 주민규 and Final OK by SGLee
Solve the system using Gauss-Jordan elimination.
Sol) The forth row is the double of the second row, so we now ignore it in this system of linear equations.
,
So .
∴ ▪
We can also solve this problem by using Sage
http://math1.skku.ac.kr/home/pub/2430/
Aug =matrix([[1,2,-3,4],[1,3,1,11],[2,5,-4,13]])
print Aug.rref()
[1 0 0 1]
[0 1 0 3]
[0 0 1 1] okay ■
2015.9.10. Solved by 김성찬, 2015.9.10. Revised by 김원경, 강민성, 2015.9.30. Refinalized by 김이곤
In the following circuit, write a linear system to find current .
[Kirchhoff's Law : ⓵ where is the current 1)
⓶ where is the voltage 2)]
, (From ⓵)
, (From ⓶)
, , , ,
⇒ where , ,
]
coefficient matrix
of )
Current : , , , . ■
Double checked by Sage http://math1.skku.ac.kr/home/math2013/343/
Aug =matrix([[1,-1,0,0,0],[0,1,-1,-1,0],[0,1,3,0,9],[0,1,0,6,9]])
print Aug.rref()
[1 0 0 0 3]
[0 1 0 0 3]
[0 0 1 0 2]
[0 0 0 1 1]
2015.10.29.Solved by 주영은 2015.10.29. Revised by 서승완 2015.10.29. Finalized by 변희성
In general, we are given a linear system with equations and unknowns.
If there are free variables, what is the number of leading variables? From this, think about the relation among the numbers of free variables, leading variables, and unknowns.
Sol.
In a homogeneous linear system with n unknowns, if the RREF of the augmented matrix has k leading 1’s, the solution set has n-k free variables. From this, we can conclude that the sum of the number of free variables and that of leading variables is a whole number of unknowns.
This problem does not need to use Sage. ■
9.12 Solved by 김도희 9.12 Revised by 2014311792 정진규 10.11 Finallized by 2014311792 정진규 2015.10.11. Final OK by SGLee
Write a linear system with 4 unknowns and 3 equations whose solution set is given below.
(here , are any real numbers)
Solution.
Using and , we can make equations.
⋯①
⋯②
Because dimension of is 2, there are 2 linearly independent equations.
⋯①+②
∴
■
9.14 Solved by 2009313441 김동욱, 9.21 Finalized by 2013312668 김이곤, 9.22 Re-finalized by 2013312668 김이곤, 10.08 Re-finalized by 2009313441 김동욱, 10.08 Final OK by SGLee
<Chapter 2 p3 NEW>
Suppose that three points pass through the parabola .
By plugging in these points, obtain three linear equations. Find coefficients by solving .
Sol)
(, , pass through the parabola)
, where ,
, ,
■
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2487
(by 2009313441 김동욱) Okay ■
2015.10.29. Solved by 주영은, 2015.10.29. Revised by 서승완
Write a linear system with four unknowns and four equations satisfying each condition below.
(a) A solution set with one unknown.
(b) A solution set with two unknowns.
Sol. (a) is an example.
Let be the matrix . By using a Sage, RREF(A) is .
Free variable is only . Therefore, this example is a solution set with one unknown.
(b) is an example.
Let B be the matrix . By using a Sage, RREF(B) is .
Free variables are . Therefore, this example is a solution set with two unknowns.
Double checked by Sage> http://math3.skku.ac.kr/home/pub/70
A = matrix(4,4,[1,-2,4,3,2,-5,1,-2,3,4,-2,5,2,-3,3,2])
print A.rref()
B = matrix(4,4,[1,1,2,-4,1,2,2,-4,1,3,2,-4,1,2,2,-4])
print B.rref()
[ 1 0 2 -4]
[ 0 1 0 0]
[ 0 0 0 0]
[ 0 0 0 0] OK ■
You can practice this code in http://sage.skku.edu/ .
2015.11.22. Solved by 강영훈 2015.11.22. Revised by 송선민 2015.11.22. Finalized by 송선민 Final OK by SGLee
● http://matrix.skku.ac.kr/LA-Lab/index.htm
● http://matrix.skku.ac.kr/knou-knowls/cla-sage-reference.htm
Indicate whether the statement is true (T) or false (F). Justify your answer.
(a) If three nonzero vectors form a linearly independent set, then each vector in the set can be expressed as a linear combination of the other two.
False
(b) The set of all linear combinations of two vectors and in is a plane.
False
(c) If u cannot be expressed as a linear combination of and , then the three vectors are linearly independent.
False
(d) A set of vectors in that contains is linearly dependent.
True
(e) If {, , } is a linearly independent set, then so is the set {, , } for every nonzero scalar .
True
Note :
(a) If three vectors are linear independent, it is impossible to make one vector with other two vector's linear combination
(b) If and are linear dependent, linear combinations of two vectors are not plane.
(c) If = (1, 0), = (0, 1), = (0, 2), we cannot make by linear combination , . However, and are linearly dependent.
(e) If a solution of is only , a solution of is . So {} are linearly independent.
2015.9.15. Solved by 강민혁 2015.9.18. Revised by 변희성 2015.11.11. Finalized by 송선민 Final OK by SGLee
When , confirm the following. .
i)
ii)
■
Sage) http://math1.skku.ac.kr/home/math2013/297/
■
2015.9.28. solved by 박원기 2015.09.29. revised by 김성찬 2015.09.29 Finalized by 박원기 2015.11.11. Refinalized by 송선민 Final OK by SGLee
When , confirm that and that .
,
but ■
Sage) http://math3.skku.ac.kr/home/pub/20 김성찬
A=matrix(2, 2, [-2, 3, 2, -3])
B=matrix(2, 2, [-1, 3, 2, 0])
C=matrix(2, 2, [-4, -3, 0, -4])
print "AB="
print A*B
print "AC="
print A*C
AB=
[ 8 -6]
[-8 6]
AC=
[ 8 -6]
[-8 6] ■
You can practice this code in http://sage.skku.edu/ .
2015.9.26. Solved by 서승완 2015.9.27. Revised by 배성준 2015.9.27. Finalized by 변희성 Final OK by SGLee
When , compute the following. .
∴ Answer is ■
Sage) http://math1.skku.ac.kr/home/pub/2478 by 배성준
A=matrix(2,2,[1,3,4,-1])
I=identity_matrix(2)
print 3*A^3-2*A^2+5*A-4*I
[ 14 132]
[176 –74] OK ■
You can practice this code in http://sage.skku.edu/ .
2015.09.19. solved by 백지원 2015.09.19. finalized by 이종현 2015.09.30 Refinalized by 이종현 Final OK by SGLee
Show that is the inverse of . And confirm that .
====
∴=
=
=
==
∴= ■
2015.9.26. Solved by 서승완 2015.9.27. Revised by 배성준 2015.9.27. Finalized by 변희성 2015.9.30. Refinalized by TA Lee Final OK by SGLee
If , show that .
∴ If , then . ■
Solved by 주영은, 2015.10.14. 김원경, 2015.10.14. Refinalized by 서승완,이나을, Final OK by SGLee
Find a elementary matrix corresponding to each elementary operation.
(1)
(2)
(3)
(Elementary matrix)
(1) : Interchange the 2nd and the 3rd rows on
(2) : Multiply the 2nd row by 2.
(3) : Add –2 times the 1st row to the 3rd row. ■
Double checked by Sage) http://math3.skku.ac.kr/home/pub/55 by 주영은
#elementary_matrix=matrix([[1,0,0], [0,1,0],[0,0,1]]) E1=elementary_matrix(3, row1=1, row2=2) E2=elementary_matrix(3, row1=1, scale=2) E3=elementary_matrix(3, row1=2, row2=0, scale=-2) print "E1 =" print E1 print "E2 =" print E2 print "E3 =" print E3 |
E1 = [1 0 0] [0 0 1] [0 1 0] E2 = [1 0 0] [0 2 0] [0 0 1] E3 = [ 1 0 0] [ 0 1 0] [-2 0 1] |
Note) Sage의 Index는 0부터 시작함을 주의 ■
You can practice this code in http://sage.skku.edu/ .
2015.11.8. Solved by 송선민 2015.11.11. Finalized by 송선민 Final OK by SGLee
Using elementary operations, find the inverse of the following matrix.
(1) (2)
(1) = → → =. =
(2) =
→ → = .
. ■
2015. 9. 27 Solved by 이나을 2015.11.11. Finalized by 송선민 Final OK by SGLee
Let and be any matrix.
(1) What is and confirm how affects on .
(2) What is and confirm how affects on .
Let = ,
(1) = =
affects on the 3rd row of .
(2) = =
affects on the 1st column of . ■
2015.9.17. solved by 강민혁 2015.9.18. Revised by 김성찬 2015.09.23. Finalized by 김성찬 2015.09.30.
Refinalized by 김성찬 Final OK by SGLee
Determine if is a subspace of .
Show 1) is closed under the addition.
2) is closed under the scalar multiplication.
1)
2)
Therefore, is not a subspace of . ■
Solved by 2013313863 강영훈 2015.9.18. Revised by 김성찬 2015.9.23.
Finalized by 김성찬 2015.09.30. Refinalized by 김성찬 Final OK by SGLee
Determine if is a subspace of .
Show 1) is closed under the addition.
2) is closed under the scalar multiplication.
1)
2)
Therefore, is a subspace of . ■
Find a vector equation and a parameterized equation of the subspace spanned by the following vectors.
(a) ,
(b) ,
(a) , , where , in ℝ.
(b) , , , , . ■
2015.09.18. Solved by 주민규, 김성찬, 변희성, 이반석, 변희성
2015.11.15. Refinalized by 송선민, Final Ok by SGLee
Give a solution by finding the inverse of the coefficient matrix of the system.
Set the coefficient matrix .
Use ERO to get :
Ans) = ■
Sage ) inverse and solution
http://math1.skku.ac.kr/home/pub/2456, http://math3.skku.ac.kr/home/pub/8/
A=matrix(3, 3, [3, 0, -1, 3, 4, -2, 3, 5, -2])
b=vector([1, 1, 2])
print A.inverse()
print "x=", A.inverse()*b # 역행렬을 이용한 연립방정식의 해 구하기
print "x=", A.solve_right(b) # .solve_right( )를 이용하여 구할 수도 있다.
You can practice this code in http://sage.skku.edu/ .
2015.9.18. solved by 강민혁 2015.9.19 revised by 이종현 2015.11.11. Finalized by 송선민 Final OK by SGLee
Determine if the homogeneous system has a nontrivial solutoin.
Let = : Augmented matrix
=
: RREF()
(3, 0, -2, 1) is one of solutions for the given homogeneous system of equations.
Therefore the system has a non trivial solution. ■
2015.9.26. Solved by 서승완, 2015.9.27. Revised by 배성준, 2015.9.27. Finalized by 남기현, 2015.10.12. Refinalized by 박시현, Final OK by SGLee
Check if the following matrix is invertible. If so, find its inverse by using a property of special matrices.
The matrix is a diagonal matrix.
Therefore .
Let , , .
, , .
=>
=>
∴ The inverse matrix of is . ■
Sage) http://math1.skku.ac.kr/home/pub/2479 by 배성준
A=matrix(3,3,[2,0,0,0,-5,0,0,0,3])
print A.inverse()
[ 1/2 0 0]
[ 0 -1/5 0]
[ 0 0 1/3] ■
You can practice this code in http://sage.skku.edu/ .
2015.10.14.Solved BY 주영은, 서승완, 김원경, 이반석, 이나을, 송선민, Final OK by SGLee
Find the product by using a property of special matrices.
, : diagonal matrices
1) = : , ,
was multiplied on the left.
2) = : ,
was multiplied on the right.
∴ The answer is . ■
Double checked by Sage)
http://math3.skku.ac.kr/home/pub/56 by. 주영은
A=matrix([[2,0,0], [0,-1/2,0], [0,0,-5]])
B=matrix([[2,4], [-4,2], [3,2]])
C=matrix([[2,0], [0,-1/2]])
print A*B*C
[ 8 -4]
[ 4 1/2]
[-30 5] : OK
http://math3.skku.ac.kr/home/pub/58 by 김원경 -(Use Diagonal)
A=diagonal_matrix([2,-1/2,-5])
B=matrix([[2,4], [-4,2], [3,2]])
C=diagonal_matrix([2,-1/2])
print A*B*C
[ 8 -4]
[ 4 1/2]
[-30 5] : OK ■
You can practice this code in http://sage.skku.edu/ .
2015.9.17. Solved by 김성찬 2015.9.23. Finalized by 김성찬 2015.09.30. Refinalized by 김성찬 Final OK by SGLee
Determine so that is skew-symmetric matrix.
The matrix is a skew-symmetric matrix then and .
The answer is
. ■
If satisfies and , show that can be expressed as follows.
What is the value of ?
=>
■
2015.11.22. Solved by 강영훈 2015.11.22. Revised by 송선민 2015.11.22. Finalized by 송선민 Final OK by SGLee
Let be a square matrix. Explain why the following hold.
(1) If contains a row or a column consisting of 0's, is not invertible.
(2) If contains the same rows or columns, is not invertible.
(3) If contains a row or column which is a scalar multiple of another row or column of .
(1) is not invertible. det = 0
det
( contain a row or a column of all zeros)
is not invertible.
(2) If a matrix has , which are , we can make a new matrix
which take on . Because the matrix has a row or a column
consisting of 0's and det=det, det=det=0. So, is not invertible.
is not invertible.
(3) If a matrix has , which are ( is constant), we can make
a new matrix which take on . Because the matrix has a
row or a column consisting of 0's and det = det, det = det = 0. So, is not
invertible.
is not invertible. ■
2015.10.13. Solved by 정범진 2015.10.13. Revised by 김성찬 2015.11.11. Finalized by 송선민 Final OK by SGLee
Let be an square matrix. Discuss what condition is need to have .
If there is an inverse matrix ,
So there must be an inverse matrix of the matrix . ■
Note : 가 invertible이 아니면 성립하지 않을 수 있다.
2015.11.22. Solved by 강영훈 2015.11.22. Revised by 송선민 2015.11.22. Finalized by 송선민 Final OK by SGLee
Find matrices , and explain the relation with ERO.
■
2015.10. solved by 강영훈 2015.10.12. revised by 신영준 2015.10.13. finalized by 김선호 2015.11.11. Refinalized by 송선민
Decide if the following 4 vectors are linearly independent.
, , ,
This implies for some .
Ex)
Ans) are linearly dependent. ■
Checked by Sage
http://math1.skku.ac.kr/home/pub/2491
A=matrix([[4,2,6,4],[-5,-2,-3,-1],[2,1,3,5],[6,3,9,6]])
print A.rref()
[ 1 0 -3 0]
[ 0 1 9 0]
[ 0 0 0 1]
[ 0 0 0 0] ■
You can practice this code in http://sage.skku.edu/ .
2015.10.15. Solved by 정다산 2015.10.16. Revised by 정다산 2015.11.11 Revised by 송선민 Finalized and Final OK by SGLee
If and have a solution, prove that has a solution.
If and have a solution, prove that has a solution.
Let and be solutions of and respectively.
=> and
=>
=> is a solution of
Therefore , if both and have a solution, then has a solution. ■
2015.11.22. Solved by 박지환 2015.11.22. Revised by 송선민 2015.11.22. Finalized by 송선민 Final OK by SGLee
Suppose is an invertible matrix of order . If in is orthogonal to every row of , what is ? Justify your answer.
For in is orthogonal to every row of ,
=0,=0=0 =0
Null()
is a solution of . ■
2015.11.22. Solved by 박지환 2015.11.22. Revised by 송선민 2015.11.22. Finalized by 송선민 Final OK by SGLee
Prove that a necessary and sufficient condition for a diagonal matrix to be invertible is that there is no zero entry in the main diagonal.
det= 0 for all () ■
If is invertible and symmetric, so is .
, and .
=> => => is symmetric. ■
2015.09.22. Solved by 주영은 2015.09.23. Finalized by 김원경 2015.09.30.
Refinalized by 김원경 Final OK by SGLee
Is permutation of even or odd?
Sol)
The number of inversions for 2 is 2. The number of inversions for 0 is 0, for 1 is 0, for 3 is 0, for 5 is 0, for 10 is 2, for 8 is 1 and 7 is the last index.
Hence, the total sum is
Answer) It is an odd permutation. ■
Double Checked by Sage)
http://math1.skku.ac.kr/home/pub/2470
Permutation([2,0,1,3,5,10,8,7]).inversions()
[[0, 1], [0, 2], [5, 6], [5, 7], [6, 7]] (5 inversions)
■
2015.9.27.solved by 변희성, 2015.9.28. Revised by 서승완, 2015.09.29.
Finalized by 김성찬, 2015.09.30. Refinalized by 김성찬, Final OK by SGLee
(New) Find the following determinants.
(1).
Sol)
()
()
()
(By Theorem 4.1.9)
So =-1.
Double checked by Sage : http://math3.skku.ac.kr/home/pub/19 (by SC.Kim)
A=matrix(5,5,[1,0,1,0,1,1,1,1,1,1,1,1,0,-1,1,0,1,0,1,1,1,0,1,-1,1])
print "det(A)=",; print A.det()
det(A)= -1 ■
[Note: 문제에서는 5차 정사각행렬이 주어졌기 때문에 어느 정도의 과정을 거치면
금방 determinant를 구할 수 있지만 차수가 점점 올라갈수록 복잡해진다. Sage를 이용한다면 어떠한 행렬에 관하여도 위와 같은 풀이를 이용하여 해결할 수 있다는 것을 알 수 있다. ]
2015. 11. 08. Solved by 강민성 2015. 11. 09. Revised by 서한길 2015. 11. 11.
Finalized by 김이곤 Final OK by SGLee
Let be matrix and , find the followings.
(1) (2) (3) (4)
Sol)
(1)
(2)
(3)
(4)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2503 (by MS.Kang)
A=matrix(QQ,3,3,[1,0,0,0,1,0,0,0,-4])
print A
print A.det()
print (A*A).det() # (1)
print (A.inverse()).det() # (2)
print (2*A).det() # (3)
print ((2*A).inverse()).det() # (4)
[ 1 0 0]
[ 0 1 0]
[ 0 0 -4]
-4
16 #
-1/4 #
-32 #
-1/32 # ■
[ Note : Determinant의 성질을 알게 되면 다양한 변형된 행렬의 Determinant를 복잡한 계산 없이 아주 손쉽게 풀 수 있다는 것을 보여주는 대표적인 문제이다. ]
2015. 11. 08. Solved by 강민성, 2015. 11. 11. Revised by 서한길, 2015. 11. 11. Finalized by 김이곤, Final OK by SGLee
For given matrices.
,
(1) show .
(2) show .
Sol) (1) ,
(2)
Double Checked by Sage : http://math1.skku.ac.kr/home/pub/2513/ (by LG.Kim)
(1)
A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])
B=matrix(QQ, 3, 3, [1, 0, 3, 0, 5, 0, 2, 0, 4])
print "det(A)=", A.det()
print "det(B)=", B.det()
det(A)= -10
det(B)= -10
(2)
A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])
B=matrix(QQ, 3, 3, [1, 2, 0, 3, 4, 0, 0, 0, 5])
C=A*B
print "det(AB)=", C.det()
print "det(A)*det(B)=", A.det()*B.det()
det(AB)= 100
det(A)*det(B)= 100 ■
[ Note 1 :
인 특수한 행렬에 대해서 를 구해보면,
,
또한, 위의 문제 (2) 에서 행렬 는 행렬 의 2번째-3번째 열을 바꾸고, 그 다음 2번째-3번째 행을 바꾼 형태이므로 Theorem 4.1.3 에 의해 이 성립한다.
실제로, Sage 를 통해 확인해보면; http://math1.skku.ac.kr/home/pub/2514 (by LG.Kim)
A=matrix(QQ, 3, 3, [1, 0, 2, 0, 5, 0, 3, 0, 4])
B=matrix(QQ, 3, 3, [1, 2, 0, 3, 4, 0, 0, 0, 5])
print A.det()
print B.det()
-10
-10
성립한다는 것을 확인할 수 있다. ]
[ Note 2 :
Determinant 의 성질이 True 임을 알 수 있는 대표적인 문제이다. 실제로 matrix 일 때에도 성립한다는 것을 보일 때
Determinant 의 성질이 완벽히 참이라는 것이 증명이 되지만 여기서는 그 경우까지 고려하면 약간 복잡해지기 때문에 이 페이지에 싣지는 않을 것이다.
그렇지만 일반적인 matrix 에 대하여 가 성립하며 이는 upper triangular matrix 또는 diagonal matrix 등 특수한 matrix 의 성질을 이용하여 증명할 수 있다. ]
2015.9.28. Solved by 서승완 2015.9.28. Revised by 변희성 2015.9.29. Finalized by 이승열
2015.10.9. Refinalized by 신영준 Final OK by SGLee
For which and , the given matrix is invertible?
Sol)
invertible => .
The matrix is a lower triangular matrix. By Theorem 4.1.9,
.
∴ Answer ; ■
2015. 11. 08. Solved by 강민성, 2015. 11. 10. Revised by 서한길, 2015. 11. 11. Finalized by 김이곤, Final OK by SGLee
For given matrices,
(1) show .
(2) show .
(3) show .
Sol)
(1)
(2)
(3)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2515/ (by LG.Kim)
(1)
A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])
B=matrix(QQ, 3, 3, [0, 1, 2, -1, 4, 0, -2, 1, -4])
print A.det()
print B.det()
10
10
(2)
A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])
B=matrix(QQ, 3, 3, [3, 6, 0, 0, 4, 1, 0, 1, 5])
C=A*B
print A.det()*B.det()
print C.det()
570
570
(3)
A=matrix(QQ, 3, 3, [0, -1, -2, 1, 4, 1, 2, 0, -4])
B=A.inverse()
print 1/A.det()
print B.det()
1/10
1/10 ■
You can practice this code in http://sage.skku.edu/ .
2015. 11. 08 Solved by 강민성 2015. 11. 10. Revised by 서한길 2015. 11. 11
Finalized by 김이곤 Final OK by SGLee
Find all cofactors of the following matrices.
(1) (2)
Sol)
(1) , ,
, ,
, ,
(2) (Using Sage)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2516 (by LG.Kim)
(1)
A=matrix(QQ, 3, 3, [1, 1, 5, 3, -6, 9, 2, 6, 2])
B=A.adjoint()
C=B.transpose()
print C
[-66 12 30]
[ 28 -8 -4]
[ 39 6 -9]
(2)
A=matrix(QQ, 5, 5, [1, 2, 3, 4, 5, 0, 1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 2, 3, 4, -4, 2, -3, 1, 5])
B=A.adjoint()
C=B.transpose()
print C
[-18 14 22 -14 -4]
[ 36 -28 -44 28 8]
[-18 14 22 -14 -4]
[ 18 -14 -22 14 4]
[ 0 0 0 0 0] ■
[ Note :
위 문제에서와 같이 (1)번 같은 경우는 손으로 쉽게 해결할 수 있지만, (2)번과 같이 행렬의 차원이 점점 커질 때 손으로 풀기 힘들어지고 복잡해진다. 그래서 Sage Code를 이용하여 문제를 쉽게 풀 수 있었다. ]
You can practice this code in http://sage.skku.edu/ .
2015.09.21. solved by 강민혁 2015.09.22. Revised by 김성찬 2015.09.22. Finalized by 김원경 2015.09.30. Refinalized by 김원경 Final OK by SGLee
Find the determinant of the matrix by cofactor expansion.
Sol)
Use cofactor expansion on the 1 column .
⇒ .
Use cofactor expansion on the 3rd column .
⇒
The answer is .
Double checked by Sage : http://math3.skku.ac.kr/home/pub/9 (by SC.Kim)
A=matrix(QQ, 4, 4, [4, 1, 0, 2, -1, 3, 5, 1, 0, 1, 4, 0, 2, -1, -1, 1])
print "det(A)=", A.det()
det(A)= 30 ■
You can practice this code in http://sage.skku.edu/ .
2015. 11. 08. Solved by 강민성 2015. 11. 10. Revised by 서한길 2015. 11. 11. Finalized by 김이곤 Final OK by SGLee
Find the adjoint matrix of the matrix form (Problem 8).
Sol) Checked by Sage : http://math1.skku.ac.kr/home/pub/2498/ (by MS.Kang)
A=matrix(QQ,4,4,[4,1,0,2,-1,3,5,1,0,1,4,0,2,-1,-1,1]);
print A.adjoint()
[ 10 -10 10 -10]
[ 12 0 -6 -24]
[ -3 0 9 6]
[-11 20 -17 32] ■
You can practice this code in http://sage.skku.edu/ .
2015.09.29 Solved by 백지원, 이종현, 이승열, 이승열, 박시현, Final OK by SGLee
Find the inverse matrix of the given matrix by cofactor expansion.
(1) (2)
Solution) (1) by cofactor expansion
(2) by cofactor expansion
= ,
==
Double checked by Sage)
(1) http://math1.skku.ac.kr/home/pub/2483 (by JW.Baek)
A=matrix([[1, 0, 1], [-1, 3, 0], [1, 0, 2]]) dA=A.det() adjA=A.adjoint() print "inverse of A = (1/dA)*adjA =" print (1/dA)*adjA |
inverse of A = (1/dA)*adjA = [ 2 0 -1] [ 2/3 1/3 -1/3] [ -1 0 1] |
(2) http://math3.skku.ac.kr/home/pub/25 (by SY.Lee)
A=matrix(5,5,[1,0,1,3,5,-1,3,0,7,2,1,0,2,1,8,2,-4,0,0,3,-8,9,2,5,4]) dA=A.det() adjA=A.adjoint() print "inverse of A = (1/dA)*adjA =" print (1/dA)*adjA |
inverse of A = (1/dA)*adjA = [ -18/133 23/133 2/7 -48/133 -29/133] [ -30/19 13/19 1 -4/19 -4/19] [ 999/133 -412/133 -27/7 -129/133 80/133] [ 164/133 -47/133 -5/7 -6/133 13/133] [-268/133 106/133 8/7 39/133 -18/133] |
2015.09.29. Solved by 이종현 2015.09.29. Resolved by 백지원 2015.09.30. Finalized by 이승열 2015.10.01. Refinalized by 이승열
2015.10.05. Refinalized by 이승열 Final OK by SGLee
Solve the systems of linear equations by using the Cramer's rule.
(1) (2)
(3) (4)
Sol)
(1) Let , , . =-101, =-60, =55, =-21
=, =, = (Cramer's Rule)
==, ==, ==
(2) Let , ,
=8, =4, =-8 , =0, =12. =, =, =, =
==, ==-1, ==0, ==
(3) Let , ,
=-4, =-4, =-4, =-8
=, =, =
==1, ==1, ==2
(4) Let , , .=3, =26, =-14 , =-7, =-12
=, =, =, =
== , == , == , ==-4
Double check by Sage)
http://math1.skku.ac.kr/home/pub/2484 (by JW.Baek)
(1)
A=matrix(3, 3, [3, -3, -2, -1, -4, 2, 5, 4, 1])
b=vector([3, 2, 1])
print "x=", A.inverse()*b
print "x=", A.solve_right(b)
x= (60/101, -55/101, 21/101)
(2) A=matrix(4, 4, [1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1])
b=vector([0, 2, 1, 1])
print "x=", A.inverse()*b
print "x=", A.solve_right(b)
x= (1/2, -1, 0, 3/2)
(3) A=matrix(3, 3, [1, 2, 1, 2, 2, 1, 1, 2, 3])
b=vector([5, 6, 9])
print "x=", A.inverse()*b
print "x=", A.solve_right(b)
x= (1, 1, 2)
(4) A=matrix(4, 4, [1, 1, 0, 0, 0, 1, 1, -2, 1, 0, 2, 1, 1, 1, 0, 1])
b=vector([4, 1, 0, 0])
print "x=", A.inverse()*b
print "x=", A.solve_right(b)
x= (26/3, -14/3, -7/3, -4) ■
2015.9.25. Solved by 이승열 2015.9.30. Finalized by 이승열 2015.10.01. Refinalized by 이승열
2015.10.05. Refinalized by 이승열 Final OK by SGLee
Solve the following problems by using the determinant of Vandermonde matrix.
(1)Find the line equation which passes through the two points and .which passes through the three points
(2) Find the coefficients ?a, b, c of parabolic equation and .
Sol)
(1)
Let the equation of the line to be .
The Vadermonde matrix is .
,
Ans:
(2)Let
Vardermonde matrix is
,
Ans:
Double check by Sage)
(1)
http://math1.skku.ac.kr/home/pub/2474 (by SY.Lee)
V=matrix(2,2,[1,-1,1,2])
v=V.inverse() 결과
y=matrix(2,1,[11,-10]) [ 4]
a=v*y [-7]
print a
(2)
http://math1.skku.ac.kr/home/pub/2475 (by SY.Lee)
V=matrix(3,3,[1,1,1,1,2,4,1,3,9]) 결과
v=V.inverse()
y=matrix(3,1,[3,3,5]) [ 5]
a=v*y [-3]
print a [ 1] ■
2015.9.25. Solved by 이승열 2015.9.30. Finalized by 이승열 2015.10.01. Refinalized by 이승열
2015.10.05. Refinalized by 이승열 Final OK by SGLee
Solve the following problems by using the determinant.
(1) The area of a parallelogram which is generated by two sides connecting the origin and each point and .
(2) The volume of parallelepiped which is generated by three vectors,
and .
Sol)
(1)
Let
(Area of parallelogram)
(2)
Let
(Area of parallelepiped)
Double check by Sage)
(1)
http://math1.skku.ac.kr/home/pub/2472 (by SY.Lee)
x1=matrix(2,1,[4,3])
x2=matrix(2,1,[7,5])
A=x1.augment(x2)
a=A.det()
print a.abs()
결과: 1
(2)
http://math1.skku.ac.kr/home/pub/2473 (by SY.Lee)
x1=matrix(3,1,[1,0,4])
x2=matrix(3,1,[0,-2,2])
x3=matrix(3,1,[3,1,-1])
A=x1.augment(x2).augment(x3)
a=A.det()
print a.abs()
결과: 24 ■
2015. 11. 08. Solved by 강민성 2015. 11. 10. Revised by 서한길 2015. 11. 11.
Finalized by 김이곤 Final OK by SGLee
Find the eigenvalues and eigenvectors of the following matrices.
(1) (2)
Sol)
(1)
,
,
⓵ If , then (From )
, where .
⓶ If . then .
, where .
(2)
By using Sage,
,
⓵ If , then
, where .
⓶ If , then
, where .
⓷ If , then
, where .
⓸ If , then
, where .
Double checked by Sage : (1) http://math1.skku.ac.kr/home/pub/2518/ (by LG.Kim)
(2) http://math1.skku.ac.kr/home/pub/2517/ (by LG.Kim)
(1)
A=matrix(QQ, 2, 2, [3, 0, -1, -2])
print A.charpoly()
x^2 - x - 6
solve(x^2 - x - 6==0, x)
[x == 3, x == -2]
print A.eigenvalues()
[3, -2]
(3*identity_matrix(2)-A).echelon_form()
[1 5]
[0 0]
(-2*identity_matrix(2)-A).echelon_form()
[1 0]
[0 0]
A.eigenvectors_right()
[(3, [(1, -1/5)], 1), (-2, [(0, 1)], 1)]
(2)
A=matrix(QQ, 4, 4, [-3, 0, -2, 8, 0, 1, 4, -2, -4, 10, -1, -2, 6, -4, -2, 3])
print A.charpoly()
x^4 - 118*x^2 - 168*x + 1485
solve(x^4 - 118*x^2 - 168*x + 1485==0, x)
[x == 11, x == -9, x == -5, x == 3]
print A.eigenvalues()
[11, 3, -5, -9]
(3*identity_matrix(4)-A).echelon_form()
[ 1 0 0 -1]
[ 0 1 0 -1]
[ 0 0 1 -1]
[ 0 0 0 0]
(11*identity_matrix(4)-A).echelon_form()
[ 1 0 0 -9/13]
[ 0 1 0 7/13]
[ 0 0 1 11/13]
[ 0 0 0 0]
(-5*identity_matrix(4)-A).echelon_form()
[ 1 0 0 7/5]
[ 0 1 0 7/5]
[ 0 0 1 -13/5]
[ 0 0 0 0]
(-9*identity_matrix(4)-A).echelon_form()
[ 1 0 0 2]
[ 0 1 0 -1]
[ 0 0 1 2]
[ 0 0 0 0]
A.eigenvectors_right()
[(11, [(1, -7/9, -11/9, 13/9)], 1), (3, [(1, 1, 1, 1)], 1), (-5, [(1, 1, -13/7, -5/7)], 1),
(-9, [(1, -1/2, 1, -1/2)], 1)] ■
[ Note :
만일 2차 정사각행렬 에 대하여 를 만족하는 , 이 존재하면
이 되어야 하므로 .
, .
, 는 위 이차방정식의 해이므로, , . ]
2015. 11. 10. Solved by 강민성 2015. 11. 11. Revised by 서한길
2015. 11. 13. Finalized by 김이곤 Final OK by SGLee
Explain why for the following matrix .
Sol)
Hence, has identical rows.
(Theorem.4.1.4) ■
2015. 11. 13. Solved by 김이곤 2015. 11. 13. Revised by 서한길 2015. 11. 13.
Revised by 강민성 2015. 11. 13 Finalized by 김이곤 Final OK by SGLee
Show that for two square matrices and , if for an invertible matrix , then .
Sol)
■
[ Note :
위 문제는 두 정사각행렬 , 에 대하여 이 성립한다는 사실을 알고 있다면 쉽게 풀 수 있는 문제 유형이다. ]
2015. 11. 14. Solved by 강민성 2015. 11. 15. Revised by 서한길 2015. 11. 18. Finalized by 김이곤 Final OK by SGLee
(New) Simplify the following determinant.
Sol)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2524 (by MS.Kang)
Use matrix . (One example)
A=matrix(QQ,3,3,[1,1,1,1,2,2,1,2,3]);
a=1
b=2
c=3
print A.det()
print a*(b-a)*(c-b)
1
1
■
[ Note : var{a, b, c}를 첫 줄에 입력하면 실제 미지수에 대해서도 임을 알 수 있다. ]
]2015.9.27. Solved by 송선민 2015.9.28. Revised by 서승완 2015.9.29. Finalized by 이승열
2015.9.30. Refinalized by 김이곤 Final OK by SGLee
Solution)
[ (From Theorem 4.2.2) ]
# multiply
⇒ #
⇒ # determinant each side
⇒ # using
⇒ # using
⇒ # devide both side
■
Checked by Sage
http://math3.skku.ac.kr/home/pub/18
A=matrix(QQ,[[4,1,0,2], [-1,3,5,1], [0,1,4,0],[2,-1,-1,1]]) B=matrix(QQ,[[1,2,3,4,5],[5,4,3,2,1],[2,1,3,5,4],[9,5,7,1,5],[2,1,1,5,1]]) C=matrix(QQ,[[4,1,0,2,6,8],[-1,3,5,1,1,5],[0,1,4,0,1,5],[2,-1,-1,1,1,5],[6,5,1,2,5,1],[1,6,8,8,7,6]]) print "det(adj A)= ",A.adjoint().det() print "(det A)^(n-1)",A.det()^(3) print "det(adj B)= ",B.adjoint().det() print "(det B)^(n-1)",B.det()^(4) print "det(adj C)= ",C.adjoint().det() print "(det C)^(n-1)",C.det()^(5) |
det(adj A)= 27000 (det A)^(n-1) 27000 det(adj B)= 429981696 (det B)^(n-1) 429981696 det(adj C)= 14808575318291015625 (det C)^(n-1) 14808575318291015625 |
Solved 김이곤, 2015. 11. 14. Revised 강민성, Finalized 서한길, Final OK by SGLee
Let be a matrix, and assume that
.
(1) Find . Which relation does this value have with ?
(2) Find .
Sol) [Tip : , ]
(1) (Using Sage)
, (From Problem p4)
(2) , then
,
(Using Sage)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2522 (by LG.Kim)
(1) adjA=matrix(QQ, 4, 4, [2, 0, 0, 0, 0, 2, 1, 0, 0, 4, 3, 2, 0, -2, -1, 2])
print adjA.determinant() # 8
P=(adjA.determinant())^(1/(4-1))
print P # 2
(2) A=P*adjA.inverse()
print A
[ 1 0 0 0]
[ 0 4 -1 1]
[ 0 -6 2 -2]
[ 0 1 0 1] ■
[ Note : 위 문제는 앞의 문제와 연관이 있는 문제이다.
앞서 문제에서 이 성립한다는 것을 보였기 때문에 이를 응용하여 간단히 를 구할 수 있었다. ]
2015. 11. 13. Solved by 서한길 2015. 11. 13. Resolved by 주영은 2015. 11. 13. Finalized by 김이곤 Final OK by SGLee
By using the Cramer's rule,
find the degree 3 polynomial which passes through the following four points.
(0,1), (1,-1), (2,-1), (3,7)
Sol) By substituting each point to the polynomial, we get four equations.
By using Cramer’s rule, (,)
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2521/ (By HK.Seo)
A=matrix(QQ,3,3,[1,1,1,8,4,2,27,9,3])
b=vector([-2,-2,6])
print A.inverse()*b
(1, -2, -1)
, , ■
2015. 11. 13. Solved by 김이곤 2015. 11. 15. Revised by 서한길 2015. 11. 16. Finalized by 강민성 Final OK by SGLee
Let the characteristic polynomial of matrix be . Find eigenvalues of matrix .
Sol)
,
, (multiple root)
, so
Eigenvalues of matrix are , . ■
[ Note :
차 정사각행렬 의 고윳값을 라고 할 때 (단, , , , ), 인 자연수 에 대하여 의 고윳값이 임을 보이자.
Ⅰ. 일 때
Ⅱ. 일 때 임이 성립한다고 가정하면,
일 때
일 때도 성립한다.
따라서 의 고윳값은 이다. ]
2015. 11. 13. Solved by 서한길, 주영은, 이반석, 강민성 2015. 11. 13. Finalized by 김이곤, Final OK by SGLee
Find the eigenspaces of , corresponding to each eigenvalue and show that they are orthogonal to each other in the plane.
Sol) ,
,
,
Also,
Hence, eigenspaces are orthogonal to each other in the plane.
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2520 (By HK.Seo)
A=matrix(QQ,2,2,[1,2,2,4])
print A.eigenvectors_right()
[(5, [(1, 2)], 1), (0, [(1, -1/2)], 1)]
■
[Note :
e-book을 참고하면 두 vector , (단, ,)에 대해서 내적한 값을 구하였는데 어차피 scalar 값을 고려할 필요가 없기 때문에 eigenvector 들의 내적 값을 구하여 이 되는지만 확인하면 된다. ]
2015. 11. 13. Solved by 서한길 2015. 11. 13. Resolved by 주영은 2015. 11. 13. Revised by 강민성 2015. 11. 13. Finalized by 김이곤 Final OK by SGLee
Find the characteristic polynomial of the following matrix. And find the roots of the polynomial by using the Sage.
Sol) The characteristic polynomial of is the determinant of .
, , ,
,
Double checked by Sage : http://math1.skku.ac.kr/home/pub/2523/ (By HK.Seo)
A=matrix(QQ,5,5,[1,1,2,1,1,1,2,3,2,1,2,3,1,2,1,1,2,2,3,1,1,1,1,1,7])
print A.charpoly()
print A.eigenvalues()
x^5 - 14*x^4 + 39*x^3 + 59*x^2 - 121*x + 31
[-1.898437229647213?, 0.3126193021869149?, 1.071200945566419?,
5.280735607688378?, 9.23388137420550?] ■
[ Note : 앞서 계속해서 강조되었듯이 이번 문제를 통하여 Sage 의 힘은 엄청나다는 것을 알 수 있다. 저 특성방정식을 구하고 난 뒤에 일반적으로 풀 수가 없다. 그래서 Sage 가 이 문제를 통하여 엄청 중요한 요소임을 알 수 있다. 저런 복잡한 문제들도 Sage Coding을 통하여 몇 분도 안 되어 풀 수 있다. ]
정수호 2008311136 차재훈 2015311247
임상훈 2009312503 정범진 2012313346
김동욱 2009313441 권순호 2012312121
호재원 2013314163
1. Introduce
본 과제의 목적은 우리가 한 학기동안 배워나간 Linear Algebra에 대한 내용들이 실제로 산업현장이나 수학이 아닌 다른 분야들에서 어떻게 활용되고 있는지에 대해 알아보기 위함이다. 현재 진행 중인 본 강좌가 기초 자연과학 수업인 것을 미루어 보면 본 강좌의 목적이 단순히 수업의 내용을 이용하여 문제를 풀고 답을 내는 것이 아니라 수업의 내용을 깊이 있게 이해하고 이후 각자의 전공을 공부하고 연구할 때 Linear Algebra를 충분히 활용함에 있다. 그러한 의미에서 본 과제가 함께 수업을 듣는 많은 학우들에게 유용한 정보와 새로운 자극이 되기를 바란다. 본 과제에 소개되고 있는 Project들은 국내외 대학들에서 Linear Algebra와 관련하여 소개되고 있는 Project들을 바탕으로 구성되었고 되도록 다양한 분야에 걸쳐 소개하고자 하였다.
2. Applications
1) Linear algebra in genetic3) (by 임상훈)
선형대수학의 이론은 자연과학과 공학, 사회과학의 여러 분야에서 사용되고 있다. 이 프로젝트는 선형대수학의 diagonalization, eigenvalues and eigenvectors 의 개념을 이용하여 세대를 거듭할수록 유전자형의 분포도를 간략하게 표현하기 위함이다. 이렇게 시작된 행렬이론을 통해 여러 현상의 문제를 1차 연립방정식으로 바꾼 후 행렬의 성질을 이용해 문제를 논리적으로 해결 하 수 있게 되었다.
Problem
어떤 농원에, 유전자형(AA, Aa, aa)을 가진 수많은 꽃들의 상대적 분포는 이다. 이 유전자형이 꽃들의 색상을 조절하고, 각 꽃들은 특성에 따라 유사한 유전자형에 의해 수정된다고 가정하자. 세대를 거듭한 유전자형을 표현하라.
Solution
Genotype vector, ( AA, Aa, aa in the n-th generation )
유사 유전자형 수정이 일어난다면, 가능한 경우는 다음과 같다.
위의 표는 다음과 같은 수식으로 나타낼 수 있다.
where, , ,
By graphing calculator, eigenvalues
, (by graphing calculator)
Thus,
초기 유전자형 분포에 따른 n 세대를 거듭했을 때 유전자형의 분포를 나타낼 수 있다. 만약 n이 무한대로 진행될 경우, 가 되므로 Aa 유전자형은 존재하지 않게 된다.
2) Fishing in George's Bank4) (by 차재훈)
이 모델은 영국의 낚시 산업에서, 물고기의 개체 수를 계산할 때 쓰였다고 한다. 어떤 한 종의 수명과 출산율을 설정해 놓은 후, 선형대수학을 이용하면 일정 시간이 지난 후의 그 종의 개체 수를 알 수 있다. 대구과의 물고기인 해덕(haddock)을 예로 들자면, 수명을 10년, 출산은 빠르면 3년 내에 할 수 있다고 설정한다. 이때, Leslie matrix를 사용할 수 있다.
Leslie matrix는 생태계에서 특정기간 동안 개체 수의 변화를 알 수 있게 하는 행렬이다. Leslie matrix는 정사각행렬이며, 몇 개의 알아놓아야 할 것이 있다.
는 살 개체의 수다.
는 살에서 살로 넘어갈 때 살아남는 비율이다.
는 생산력(출산율)인데, x살의 어머니 개체 1개가 낳는 암컷 자식의 수이다.
최대 수명이 년이면, Leslie matrix도 행렬이 된다.
가 된다. 또한 이것은 혹은 와 같이 표기 될 수 있다.
이것을 실제 해덕에 대해 조사를 하면 해덕의 Leslie matrix(1년 기준)는
가 된다. 이걸로 일정 기간 후의 해덕의 개체 수를 알 수 있게 된다.
Note.
해덕의 Leslie matrix 의 고유벡터들도 linearly independent하기 때문에 diagonalization을 이용하여 향후 몇 년 후의 개체수를 간단한 연산으로 추정해볼 수 있다.
3) Matrices in Statics and Mechanics5) (by 정범진)
이 프로젝트의 목표는 복잡하고 변수가 많은 역학 문제를 푸는 데 있어 선형대수학이 어떻게 쓰이는지 알아보는 것이다. 실제로 기계공학 등에서 다루어지는 역학은 계산의 규모가 크고 복잡한 경우가 많은데, 여기에 교재의 2장에서 배운 Linear system of equation을 적용하여 식을 구성하고 RREF를 적용하면 그 문제를 보다 쉽게 풀 수 있을 것이다. 그 예로 다음과 같은 문제를 풀어보려고 한다.
다음과 같은 구조물이 있다고 하자. 구조물은 힘 과 가 가해지는 상태에서 정적평형을 유지하고 있다. 이때 각 연결부위에 가해지는 내력(內力)을 구하고자 한다. 주어진 조건은 ,,,이고 은 항력, 는 철근의 장력을 의미한다. 이에 따라 각 연결부위에서의 내력을 , 축 방향으로 분리하여 식으로 나타내면 다음과 같다.
- 좌측 하단 연결부위의 축 방향 :
- 좌측 하단 연결부위의 축 방향 :
- 가운데 하단 연결부위의 축 방향 :
- 가운데 하단 연결부위의 축 방향 :
- 우측 하단 연결부위의 축 방향 :
- 우측 하단 연결부위의 축 방향 :
- 좌측 상단 연결부위의 축 방향 :
- 좌측 상단 연결부위의 축 방향 :
- 우측 상단 연결부위의 축 방향 :
- 우측 상단 연결부위의 축 방향 :
이를 행렬을 통해 표현하면
로 나타낼 수 있고
여기에 RREF를 하면
이렇게 각각의 항력,장력의 크기 역시 쉽게 구할 수 있다.
이와 같이 실생활에서는 variable이 많고 복잡한 문제를 접하게 되는데, 이때 선형대수학을 이용한다면 그냥 손으로 푸는 것보다 훨씬 쉽고 빠르게 답을 구할 수 있다는 것을 알 수 있다. 또한 이런 RREF와 같은 연산을 할 때 sage와 같은 연산 프로그램의 도움을 빌리는 것이 시간을 절약할 수 있을 것이다.
4) Singular Value Decomposition6) (by 정수호)
소개하고자 하는 프로젝트는 SVD를 이용한 영상이미지의 데이터 축소에 관한 것입니다. 간단히 소개하자면 SVD(Singular Value Decomposition)는 데이터 저장 및 전송 최소화에 유용한 수단으로써, 현대사회에 디지털 정보의 양을 축소시키고자 하는데 활용되어지고 있는 선형대수학의 중요 적용사례입니다. SVD는 강의 “8-6“에서 다루어지고 있으며 일반화된 역행렬, 최소 제곱해 등으로 사용되어지며 나아가 데이터 압축, 노이즈제거 등으로 활용되어 진다.
특이값 분해는 고유값 분해처럼 행렬을 대각화하는 한 방법이다. 그런데, 특이값 분해가 유용한 이유는 행렬이 정방행렬이든 아니든 관계없이 모든 행렬에 대해 적용 가능하기 때문이다.
실수공간에서 임의의 행렬에 대한 특이값분해(SVD)는 다음과 같이 정의된다.
는 를 고유값 분해해서 얻어진 직교행렬로 의 열벡터들을 의 left singular vector라 부른다. 또한 는 를 고유값 분해해서 얻어진 직교행렬로서 의 열벡터들을 의 right singular vector라 부른다. 마지막으로, 는 , 를 고유값 분해해서 나오는 고유값들의 square root를 대각원소로 하는 직사각 대각행렬로 그 대각원소들을 의 특이값이라 부른다.
우선 이미지는 이미지의 픽셀 높이 과 픽셀너비 의 행렬로 나타내어질 수 있으며 그레이스케일로써 행렬의 성분은 0(검정색)부터 1(흰색)으로 나타내어진다. 색상이 있는 이미지는 RGB (Red, Green, Blue)로 나누어 그레이스케일과 마찬가지로 0부터 1까지로 각각 나타내어진다. 이렇게 행렬로 나타내어진 이미지는 특이값 분해에 의해 표기 될 수 있다.
특이값 분해는 위의 수식과 같이 간단히 나타내어질 수 있으며 이므로 행렬 A는 초반부 몇 개의 성분의 합만으로 근사시킬 수 있다.
이미지 행렬을 위의 방법과 같이 근사하게 된다면 값이 커질수록 이미지의 질은 향상되지만, 이미지를 저장하기 위한 용량 또한 증가하게 된다. 이는 용량의 감소를 위해선 낮은 값이 선호될 것이란 것을 의미한다.
다음의 사진은 값이 1, 5, 20, 50, 164일 때의 이미지 모습과 원본의 모습을 순서대로 나열하여 비교해놓은 모습이다. 값에 따라 용량의 감소와 함께 이미지의 선명함 또한 감소하게 haddock된다. 하지만 어떠한 값에서 이미지의 선명함이 충분한지는 주관적인 판단에 의해 결정 되어지는 것이며 용량의 감소와 함께 적절한 타협점을 통해 조율할 수 있을 것이다.
5) Linear Algebra in data analysis (by 김동욱)
컴퓨터와 인터넷의 발전으로 인해 이전과는 비교할 수 없는 데이터들이 발생되고 또 디지털화 되어 저장되고 있다. 이렇게 증가한 데이터의 공급과 함께 데이터 활용과 수요도 폭발적으로 늘어나며 이제는 빅데이터나 IOT와 같은 데이터 관련 단어들이 전공자들의 전유물이 아니다. 이러한 데이터 분석 분야에서도 선형대수학이 많이 이용되고 있는데 그 이유는 기본적으로 대부분의 데이터는 Matrix형태로 관리되며 데이터의 속성을 나타내는 variables이 늘어날수록 데이터의 차원이 확장되므로 선형대수학적인 접근이 필수적이다. 본 프로젝트에서 소개한 선형대수학을 이용한 데이터 처리 기법은 PCA이다.
PCA(주성분 분석)7)
PCA는 통계학이나 데이터분석 분야에서 데이터의 차원을 축소시키는 기법 중 하나이다. PCA를 간단히 설명하자면 표준기저로 표현되어 있는 데이터들을 우리가 7장에서 학습한 기저 변환을 통해 그 데이터들을 더 잘 설명할 수 있는 새로운 기저를 찾고 projection을 이용하여 차원을 줄여나가는 기법이다.
새로운 기저는 다음과 같이 표준 기저의 선형결합으로 표현되며 데이터들의 차이를 크게 하기 위해 새로운 기저로 정의되는 데이터들의 분산이 가장 큰 기저를 정의한다.
표준 기저:
새로운 기저:
,
표준 기저의 선형 결합으로 정의된 새로운 기저는 각각 orthonormal하며 각 축의 데이터에 대한 설명력은 계수인 에 의해 결정된다. 앞서 설명한 것과 같이 새로운 기저로 정의된 데이터의 분산이 가장 큰 기저가 가장 큰 설명력을 가지므로 Optimization 기법을 사용하여 를 정의할 수 있으며 는 데이터의 Covariance matrix의 고유벡터가 되며, 각 축의 설명력은 고유 값의 크기에 따라 판단할 수 있다.
PCA는 간단한 기저 변환을 통해 데이터의 차원을 축소할 수 있어 기계학습과 데이터의 시각화 등에 널리 사용되고 있다.
6) 다중 안테나 다중 중계 시스템을 위한 가중치 대각화 기반의 저 복잡도 전력 할당 기법8) (by 권순호)
선형대수학은 모든 분야에 걸쳐 응용될 수 있다. 그 중 전자전기공학 분야에 관련된 좋은 사례를 찾을 수 있었다. 우리 주변에서 흔히 볼 수 있는 안테나의 성능을 향상시킬 수 있는 주요한 기술 중 하나가 선형대수학이다. 안테나의 전력으로 구성된 실수 대각 행렬을 전처리 필터로 사용하는데, 이 행렬은 복소수로 구성되어 있다. 하지만 이 복소수 행렬로 유효 채널을 대각화 하지 못함에서 오차가 크게 발생할 수 있다. 우리는 이 오차를 감소시키기 위하여 행렬을 대각화해서 최적 성능에 근접한 성능을 얻는 안테나 설계를 할 수 있다.
기존 다중 안테나 기반의 단일 중계 노드를 위한 전처리 기법들은 유효 채널을 대각화하기 위해서 선형대수학 수업 시간에 배운 8장의 특이값 분해(SVD) 기술을 사용하였다. 이 기술은 복잡도 높은 수학 기술 중 하나이기 때문에 이 사례에서는 새로운 개념의 “가중치 대각화”를 소개한다.
위 그림은 하나의 송신 노드(S)와 하나의 수신 노드(D) 사이에서 R개의 중계 노드가 있는 다중 안테나 시스템이다. 이 때 수신된 신호 는 다음과 같다.
여기서 행렬H=은 송신 노드와 중계 노드들 사이의 채널을 의미하고, 행렬G=은 중계 노드들과 수신 노드 사이의 채널의 의미한다.
위 식을 이용하여 시스템의 상호정보량(I)을 다음과 같이 나타낼 수 있다.
만약 송신 노드와 수신 노드의 안테나 수가 중계 노드와 중계 노드의 안테나 수보다 많다면 큰 수의 법칙에 의해 행렬과 행렬의 비대각 성분들을 무시할 수 있다. 그러면 특이값 분해(SVD)의 복잡한 기술의 적용 없이 상호정보량(I)는 다음과 같이 근사화 된다.
이것이 바로 이번 연구 사례에서 제안하는 가중치 대각화 기반의 전력 할당 기법이다.
위 행렬에서 비대각 성분의 평균값은 채널 벡터 상호간의 독립성에 의해 0이다. 반면에 행렬의 대각 성분은 각 채널 벡터의 놈(norm)으로서 항상 0 이상의 값을 갖는다. 만약 송신 노드의 안테나 수가 중계 노드와 중계 노드 안테나 수의 곱보다 크다면 큰 수의 법칙에 의해 비대각 성분의 값은 0에 가까워질 것이고, 반대로 대각 성분의 값은 증가한다. 따라서 다음과 같이 근사적으로 대각화 할 수 있다.
같은 원리에 의해 만약 수신 노드의 안테나 수가 중계 노드와 중계 노드 안테나 수의 곱보다 크다면 는 다음과 같이 근사적으로 대각화 된다.
7) 3차원 천이행렬을 대각화 하는 간단한 방법 : Fast Multipole9) (by 호재원)
(1) Introduction
선형대수학은 실생활에서 응용되는 분야가 아주 다양하다. 특히 공학적 시스템에서 데이터의 처리 및 다변수 문제의 해결에서 아주 중요한 부분을 차지하는 학문이다. 이번 프로젝트에서는 이러한 선형대수학을 실제로 응용한 사례를 찾아 그 이론과 응용 방법 등을 밝히는 것을 그 목표로 한다.
탐구하고자하는 이론은 선형대수학 수업시간에도 다루어졌던 Chapter 8에 해당하는 행렬의 대각화이다. 이 이론은 공학수학에서도 다루어지는 내용으로써 데이터의 처리에 상당한 이점을 제공한다. 데이터의 행과 열로 정리된 하나의 행렬은 개의 방정식과 개의 미지수로 이루어진 것이다. 한편 이들은 서로 Elementary row operation을 통해 동치관계의 다양한 행렬을 생성할 수 있다. 이 때 행렬의 각 성분이 단순할수록 더 적은 데이터 요구치를 달성할 수 있으며 이는 보다 효율적으로 데이터를 압축하여 활용한다는 것이다. 더하여 이 방법은 데이터의 계산에도 높은 효율을 보여준다. 일일이 행을 곱하고 더하는 것이 아니라 최소한의 성분만을 곱 또는 합에 사용함으로써 데이터 처리의 이득을 얻는 것이다. 이를 실제적으로 사용한 사례를 이 프로젝트에 싣고자 한다.
(2) Subject
대각화는 전기역학적 또는 탄성역학적 문제에 대한 적분 방정식의 해를 신속히 처리하는데 유용하다. 이 방정식들은 다양한 변수를 취급하기 마련이다. 이에 한 문제에서 얻어지는 다양한 방정식은 하나의 행렬를 구성한다. 일반적으로 곧장 문제를 해결하려하면 수많은 행렬과 벡터의 값을 계산해야하며, 컴퓨터를 사용해도 그 필요한 값만큼 대용량의 메모리를 요구한다. 제시된 연구 논문에서 소개된 이론은 하나의 좌표시스템에서 또 다른 좌표시스템으로의 천이를 기본으로 한다. 하지만 단순한 천이 행렬의 사용은 컴퓨터의 산술이나 메모리 요구치의 감소에 아무런 도움을 주지 않는다. 이런 문제점을 해결해주는 것이 바로 천이 오퍼레이터의 대각화이다. 대각화로써 각 방정식들은 훨씬 간편해지고 단순해진다. 이는 신속성과 편리함을 제공하는 것을 이 연구논문에서 확인 할 수 있다.
3. Conclusion
본 과제를 통해 Linear Algebra를 활용한 여러 예들을 살펴보았다. 위에서도 볼 수 있듯이 역학이나 생물학뿐만 아니라 컴퓨터공학이나 전자전기 분야에서도 Linear Algebra가 많이 활용되고 있다. 본 과제에는 다 실지 못했지만 이 이외에도 무수히 많은 분야에서 무수히 많이 활용되고 있다. 대부분의 분야들에서 복잡한 연산을 처리하거나 복잡한 문제를 단순화하거나 축소시키는데 Linear Algebra가 많이 활용되고 있다는 것을 확인 할 수 있었다. 이는 Linear Algebra가 갖는 힘과 의미를 잘 보여준다.
본 과제를 진행하며 Linear Algebra에 대해 더 이해하고 Linear Algebra가 가지고 있는 가능성을 알게 되었지만 아쉬운 부분들이 많다. 본 과제에 소개되고 있는 예시들이 이미 오래전에 개발된 방법들이거나 대학생들이 제시한 방법들이여서 완성도가 다소 떨어지는 면이 없지 않아 있다. 또한 본 과제가 그저 활용된 예들에 대한 소개에만 그쳐 새롭고 좀 더 의미있는 내용들로 구성되지 못한 점도 큰 아쉬움으로 남는다.
4. 후기
(1) 정수호 2008311136
이번 프로젝트 조사를 통해 선형대수학이 우리 주변 곳곳에서 활용되고 있으며 없어선 안 될 기반이 되는 부분임을 알 수 있었다. 이론을 통해 응용하여 실제적인 적용을 이끌어 내야할 공학도로써 이렇게 선형대수학이 적용되어 사용되는 사례를 보고 많은 생각을 하게 되었다. 이미지 용량의 감소뿐만 아니라 다양한 사례를 통해 이용되어지는 선형대수학과 더불어 다양하게 적용되어질 학문들을 배우는데 있어 보다 완벽하고 철저한 태도로 임해야겠다고 생각하였다.
(2) 차재훈 2015311247
선형대수학이 실생활에 많이 응용이 되는 분야라는 것을 알 수 있게 되었다. 이러한 방법을 이용하면, 생태계의 동향을 파악하는데 많은 도움이 될 것 같다.
(3) 임상훈 2009312503
선형대수학을 이용하여, 어떻게 주위의 문제를 해결하는데 쓰이는지를 알 수 있었습니다. 선형 대수학을 이용하여 주위의 business, economics, society, environment 등에 적용될 수 있고, 특히 이번 프로젝트를 조사하면서 유전학에서도 응용될 수 있다는 것을 느꼈습니다. mathematical modeling을 통해 궁극적으로 문제에 대한 더욱 깊은 이해를 이끌게 되고 그 문제와 관련된 더욱 정확한 의사 결정을 내리는 것이 가능하도록 해준다는 것을 알았습니다.
(4) 정범진 2012313346
이번 프로젝트를 진행하면서 선형대수학을 물리적 상황에 접목시키는 방법에 대해 조금이나마 깨닫게 된 것 같다. 지금까지는 물리적인 상황이 주어졌을 때 그것을 최대한 단순화하여 모델링하는 것에 집중했지만, 선형대수학을 이용하여 다변수 함수를 풀어가는 방법에도 눈을 뜨게 되었다. 이론을 실제로 적용시켜보는 훈련은 수업에서 배우기 어려운 부분인데 이렇게 프로젝트를 통해서 배울 수 있게 되어 유익하다고 느꼈다.
(5) 김동욱 2009313441
수업을 들으면서 느꼈던 부분이지만 Computer Science와 Linear Algebra가 매우 밀접한 관련이 있음을 다시 한 번 느낄 수 있었다. Linear Algebra의 많은 내용들이 언어를 통해 프로그램을 구현할 때뿐만 아니라 데이터 분석이나 데이터 관리, 통계에서도 많이 활용될 수 있을 것으로 생각된다. 또한 컴퓨터와 관련된 많은 부분에서 Linear Algebra를 활용함으로써 연산을 매우 단순화 시킬 수 있기 때문에 활용했을 때와 하지 않았을 때에 성능적인 측면에서 많은 차이를 보일 것 같다.
위에 소개한 PCA 이외에도 데이터 분석과 기계학습 분야에서는 많은 Linear Algebra 기법들을 활용되고 또 더 활용하기 위해 많은 시도들이 이어지고 있다. 그래서 이번 한 학기동안 Linear Algebra 수업을 통해 배운 내용들이 전공인 데이터 분석과 기계학습에 대한 내용들을 이해하는데 많은 도움이 되었고 앞으로 연구에 있어서도 많은 도움이 될 것으로 기대하고 있다.
(6) 권순호 2012312121
선형대수학의 전자전기공학에 대한 응용 범위는 상당히 넓다. 행렬의 대각화 부분을 응용한 기술을 이용한 안테나 성능 최적화뿐만 아니라 전자전기공학의 대표적인 또 다른 분야인 제어 시스템 설계, 의료 공학, 통신 시스템 연구 분야 등등 광범위하게 선형대수학 기술을 응용할 수 있다.
(7) 호재원 2013314163
선형대수학을 실용적으로 어디에 사용하는지는 알고 있었지만, 실제로 그 활용에 대해 알아보고 탐구한 것은 이번이 처음이다. 복잡한 데이터의 단순화에 행렬의 대각화가 유용한 방법임을 파악하였으며, 이를 탐구하다가 데이터의 압축이 이와 관련이 깊다는 것 또한 알 수 있었다. 단순하지만 그 효용이 높다는 것을 느꼈다. 강의실을 떠나 실질적 공학문제에 이용되는 것을 보고 그를 활용하는 법을 조금 더 깊게 알았다는 점에서 충분히 유익한 시간이 되었다고 느낀다.
2015.10.1. Solved by 강민혁, 2015.10.1. Revised by 김원경, 이승열, Final OK by SGLee
Chapter6 p.g224
Verify that , where , is a linear transformation and find for .
Sol)
Let .
1) (∵)
⇒
2)
⇒
Answer) is a linear transformation from . ■
for
Answer) ■
Double check by Sage)
http://math3.skku.ac.kr/home/pub/28
x, y, z = var('x y z')
h(x, y, z) = [x, x+y, x+y+z]
T = linear_transformation(QQ^3, QQ^3, h)
C = T.matrix(side='right')
x=matrix(3,1,[1,2,3])
A=C*x
print A
[1]
[3]
[6] OK ■
Solved by 2014312052 김도희 2015.10.07. Revised by 백지원 Finalized by 2014312052 김도희 2015. 10. 11 Refinalized by 이나을 Final OK by SGLee
Solution)
■
Note : 표준기저를 이용하여 문제를 구하는 것이므로 의도에 맞게 풀어야 합니다.
2015.10.7 Solved by 서승완 2015.10.7 Revised by 하지크 2015.10.7 Finalized by 이종현 2015. 10. 11 Refinalized by 이나을 Final OK by SGLee
Let a linear transformation satisfy the following conditions:
, .
(1) Evaluate .
(2) Evaluate .
Solution)
1)
2)
■
Note : Linear Transform의 성질만 이용하여 문제를 간단히 풀 수 있습니다.
2015.10.1. solved by 강민혁 2015.10.2. Revised by 이반석 2015.10.4. Revised by 강민혁 2015.10.4. Finalized by 이승열 Final OK by SGLee
Sol)
Ans: ■
Double checked by Sage)
http://math3.skku.ac.kr/home/pub/27
a=cos(2*pi/3)
b=sin(2*pi/3)
c=sin(2*pi/3)
d=-cos(2*pi/3)
A=matrix(2,2,[a,b,c,d])
v=matrix(2,1,[3,1])
T=A*v
print v
print T
결과:
[3]
[1]
[1/2*sqrt(3) - 3/2]
[3/2*sqrt(3) + 1/2] ■
2015.10.07 Solved by 서승완 2015.10.7 Revised by 하지크 2015.10.7 Finalized by 이종현 2015.10.8. Refinalized by 이승열 2015. 10. 11 Refinalized by 이나을 Final OK by SGLee
Check whether the given matrix is an orthogonal matrix. If that is the case, find the inverse matrix.
Solution)
===
Similarly
∴ = : ( is an orthogonal matrix), =
■
Double check by Sage)http://math3.skku.ac.kr/home/pub/35
■
2015.10.12. solved by 주민규 2015.10.12. revised by 이반석 215.10.12. finalized by 백지원 2015.10.13. refinalized by 신영준 Final Ok by SGLee
Problem 6
For each given linear transformation, find the kernel and range. Also determine whether it is bijective or not.
(1)
(2)
Sol)
(1) 선형변환 의 표준행렬 :
⇒ 는 가역행렬이다 의 행벡터, 열벡터 모두 일차독립이다 Im.
=0 이다 는 단사이다. Im이다 는 전사이다.
, Im , bijective ■
(2) 선형변환 의 표준행렬 :
는 비가역 행렬이다.
선형변환 를 하면 평면위의 모든 점들이 () 으로 이동한다. Im ().
=0 이다 는 단사이다. Im()이다 는 전사가 아니다.
, Im , injective ■
2015.10.4. Solved by 이승열 2015.10.5. Finalized by 박시현 2015.10.6. Refinalized by 이승열 Final Okay by SGLee
Chapter6 p.g225
Let and are defined as follows:
.
(1) Find the standard matrix for each and .
(2) Find the standard matrix for each and .
Sol)
(1)
■
(2)
■
Double check by Sage) http://math3.skku.ac.kr/home/pub/29
x,y,z=var('x y z') A(x,y,z)=(4*x,-2*x+y,-x-3*y) a(x,y,z)=(x+2*y,-z,4*x-z) T=linear_transformation(QQ^3, QQ^3,A) t=linear_transformation(QQ^3, QQ^3,a) C = T.matrix(side='right') c = t.matrix(side='right') print "[T1]=" print C print "[T2]=" print c print "[T2*T1]=" print C*c print "[T1*T2]=" print c*C |
[T1]= [ 4 0 0] [-2 1 0] [-1 -3 0] [T2]= [ 1 2 0] [ 0 0 -1] [ 4 0 -1] [T2*T1]= [ 4 8 0] [-2 -4 -1] [-1 -2 3] [T1*T2]= [ 0 2 0] [ 1 3 0] [17 3 0] |
Comment : Ch.6의 2번문제와 마찬가지로 선형변환 ,는 3차원이므로 가장 간단한 단위직교벡터를 이용하여 표준행렬을 찾았고, Sage의 계산결과와 같음을 확인 할 수 있었다.
2015.10.05. solved by 김준현 2015.10.05. revised by 주민규 2015.10.05. finalized by 주민규 2015.10.06. refinalized by 이승열 Final OK by SGLee
Chapter6 p225
Sol)
■
Double check by Sage)
http://math3.skku.ac.kr/home/pub/31
Input
x_1,x_2,z_1,z_2=var('x_1,x_2, z_1,z_2')
T(x_1,x_2)=(x_1+2*x_2,x_2)
S(z_1,z_2)=(z_1,(-1)*z_1+z_2)
x(x_1,x_2)=(x_1,x_2)
A=linear_transformation(QQ^2,QQ^2,T)
B=linear_transformation(QQ^2,QQ^2,S)
C= A.matrix(side='right')
c= B.matrix(side='right')
print '(S dot T)(x)='
print (c*C)*(x)
Output
(S dot T)(x)=
(x_1, x_2) |--> (x_1 + 2*x_2, -x_1 - x_2) ■
2015.11.08. Solved by. 이반석 2015.11.08. Revised by H.Haziq 2015.11.08. FINAL OK by SGLee
Answer the following questions.
(1) Find the dimension of the null space of the following matrix by using the Sage.
(2) Let be a linear transformation corresponding to the above matrix . Determine whether is in the range of by using the Sage.
Sol.
(1)
■
(2)
is a linear combination of Columns of A.
is in the range of . ■
Note. 로 이뤄진 선형변환의 치역에 벡터 가 있으려면, 벡터 가 행렬 의 열벡터들의 일차결합으로 이루어져야 한다.
2015.11.08. Solved by 이반석 2015.11.08. Revised by H.Haziq 2015.11.09. Final OK by SGLee
Let . Find by using the Sage.
Sol) http://math3.skku.ac.kr/home/math2013/75/ by. 이반석
[cos(t) -sin(t) -x0*cos(t)+y0*sin(t) + x0]
[sin(t) cos(t) -x0*sin(t) -y0*cos(t) + y0]
[0 0 1]
Ans) [x*cos(t) - x0*cos(t) - y*sin(t) + y0*sin(t) + x0]
[x*sin(t) - x0*sin(t) + y*cos(t) - y0*cos(t) + y0]
[ 1]
Note) (x0, y0, 1)을 기준으로 x축과 y축 방향으로 각각 , 만큼 이동하였다.
이동한 값을 제곱해서 더하면 이 된다. (x, y, 1)는 theta 값에 따라서 (x0, y0, 1)을 기준으로 두 점 사이의 거리를 반지름 으로 하는 원 위의 점이 된다. ■
2015-10-22 solved by 강민혁 2015-10-23-finalized by 백지원 2015-10-26-refinallized by 김민기 2015-10-27-
refinalized by 강민성 2015-11-09-Refinalized by SC.Kim 2015-11-09-Refinalized by 이승열 2015-11-15- Refinalized by WK.Kim Final OK by SGLee
(original) Use determinant to check if the following vectors are linearly independent:
, ,
1) Solution
Therefore, is linearly independent. ■
2) Double check by Sage
http://math3.skku.ac.kr/home/pub/63 by 강민성
V=RR^3;
x1=vector([1, 1, -3])
x2=vector([0, 2, 1])
x3=vector([0, -1, 0])
A=column_matrix([x1, x2, x3]); # x1, x2, x3를 열벡터로 하는 행렬 생성
S=[x1,x2,x3]
print A.det()
print V.linear_dependence(S)
ans)
1 : not zero ok.
[] : L.I. ok. ■
2015.10.30. Solved by 서승완 2015.10.30. Finalized by 남기현 2015.11.3. Refinalized by 김원경 2015.11.8. Refinalized by 이승열 Final Ok by SGLee
LA Chapter 7 Exercises 1(new)
Use determinant to check whether the following vectors are linearly independent or not:
, , ,
Solution.
By Theorem 7.1.1. the following three vectors in
, , ,
are linearly independent because . ■
Sage http://math3.skku.ac.kr/home/pub/80 by 김원경
x1=vector([1,1,-3,2])
x2=vector([0,2,1,3])
x3=vector([0,-1,0,-4])
x4=vector([0,-1,0,5])
A=column_matrix([x1,x2,x3,x4])
print A.det()
9 ok. ■
2015.10.27. Solved by 김성찬 2015.10.29. Revised by 서승완 2015.10.30. Finalized by 정진규 2015.11.9. Refinalized by WK.Kim 2015.11.10. Refinalized by 이승열 2015.11.15. Refinalized by WK.Kim Final OK by SGLee
(NEW) Use determinant to check if the following vectors are linearly independent:
, ,
Sol)
Let a matrix where ’s are column vectors.()
Then
So the vectors are linearly independent. (Theorem 7.1.1) ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/66 by SC.Kim
Input
x1=vector([2, 1, -1])
x2=vector([3, 5, 0])
x3=vector([1, 0, 1])
A=column_matrix([x1, x2, x3])
print "det(A)=",; print A.det()
Output
det(A)= 12 : ok ■
2015.10.27. Solved by 김성찬 2015.10.29. Revised by 서승완 2015.10.30. Finallized by 정진규 2015.11.09. Refinalized by WK.Kim 2015.11.10. Refinalized by 이승열 2015.11.16. Refinalized by SC.Kim Final OK by SGLee
(New) Determine if the given set is a basis for .
(1)
(2)
Sol)
(1)
.
Hence is linearly independent. So the is a basis for . (By Theorem 7.1.4) ■
(2)
.
Hence is linearly dependent. So the is not a basis for . (By Theorem 7.1.4) ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/67 by SC.Kim
Input
A=matrix(QQ, 3, 3, [2, 6, 0, 0, 3, -5, 1, 5, 0])
B=matrix(QQ, 3, 3, [0, 2, 1, 2, 4, 3, 3, 1, 2])
print "det(A)=",; print A.det()
print "det(B)=",; print B.det()
Output
det(A)= 20
det(B)= 0 : OK ■
2015.10.30. Solved by 서승완 2015.10.30. Revised by 남기현 2015.10.31. Revised by 이반석 2015.11.02. Finalized by 백지원 Final OK by SGLee
Determine if the given set is a basis for .
(1)
Solution)
To show that is a basis for , we need to show that is linearly independent and it spans .
(1) The determinant of is .
is linearly independent.
(2) We will now show that spans .
Let be the vector in .
, by (1) ,
Indeed, the determinant of coefficient matrix of the linear system is not 0, so it is invertible. so the linear system has a solution.
spans
is a basis of ■
Sage http://math3.skku.ac.kr/home/math2013/57/
■
2015.10.27. Solved by 김성찬 2015.10.28. Revised by 김이곤 2015.10.29. Finalized by 김이곤 Final OK by SGLee
Find two different bases for the subspace of described by the equation .
Sol)
Let .
From the equation .
Let , , then .
, and is another basis. ■
[Note : , 가 주어진 방정식의 해이면 역시 주어진 방정식의 해가 된다. 위의 문제에서 로 나타낼 수 있으므로 역시 기저가 된다.]
2015.10.30. Solved by 서승완 2015.10.30. Finalized by 남기현 2015.11.4. Refinalized by 김원경 2015.11.8. Refinalized by 이승열 2015.11.16. Refinalized by SC.Kim
Final OK by SGLee
LA Chapter 7 Exercises 3(new)
Find the basis for the subspace of spanned by solutions of the equation .
Sol)
Let , where and in ℝ.
.
A basis of the null space of : ■
Note)
From that homogeneous linear system of equations, coefficient matrix , and the general solution are obtained.
2015.10.28. Solved By 김이곤 2015.10.28. Revised By 김이곤 Finalized and Final OK by SGLee
Given a homogeneous linear system, find a basis and the dimension of its corresponding solution space.
(1)
(2)
(Hint: http://math1.skku.ac.kr/home/pub/548/)
Sol) (1) , , ,
Let and where ℝ, then .
, nullity().
(2) , (Using Sage)
, ,
Let , , where ℝ, then
, nullity(). ■
(Double Checked by Sage :
http://math1.skku.ac.kr/home/math2013/397/)
(1)
A=matrix(2,4,[[4,-2,1,2],[1,1,1,-1]])
A.rref()
print A.rref()
A.rank()
A.right_nullity()
[ 1 0 1/2 0]
[ 0 1 1/2 -1]
2 OK
(2)
A=matrix(4,6,[[1,3,-2,0,2,0],[2,6,-5,-2,4,-3],[0,0,5,10,0,15],[2,6,0,8,4,18]])
A.rref()
print A.rref()
A.rank()
A.right_nullity()
[1 3 0 4 2 0]
[0 0 1 2 0 0]
[0 0 0 0 0 1]
[0 0 0 0 0 0]
3 OK ■
Note : RREF를 구한 후 Free variable을 이용하여, 해집합을 구하고, 해집합의 기저를 구하면, 바로 nullity() 를 구할 수 있었다.
Solved by 서승완 Finalized by 남기현 Refinalized by 변희성 Refinalized by SC.Kim
Final OK by SGLee
Given matrix , find a basis for its null space and nullity().
.
Sol)
The RREF of the augmented matrix for is
Hence the general solution is
.
Therefore, a basis and the dimension of the null space of is
and nullity() = 1. ■
Double checked by Sage)
http://math3.skku.ac.kr/home/pub/111 by. SC.Kim
A=matrix(ZZ,6,5,[2,2,-1,0,1,-1,-1,2,-3,1,1,1,-2,0,-1,0,0,1,1,1,0,0,0,1,1,0,0,1,1,0])
print "RREF(A)="
print A.rref()
print A.right_kernel()
print "nullity(A)=",; print A.right_nullity()
RREF(A)=
[1 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]
[0 0 0 0 0]
[0 0 0 0 0]
Free module of degree 5 and rank 1 over Integer Ring
Echelon basis matrix:
[ 1 -1 0 0 0] : OK
nullity(A)= 1 : OK ■
2015.10.27. Solved by 김성찬 2015.10.29. Finalized by 주민규 2015.11.04. Final O.K by S.G Lee
(NEW) Given matrix , find a basis for its null space and nullity().
.
Sol)
Find RREF of (Use Sage.)
RREF(A)=.
Ans) is a basis. Dimension of the null space of is nullity() = 1. ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/68 by SC.Kim
Input
A=matrix(ZZ,6,5,[3,-9,3,2,1,1,-3,2,-3,2,0,0,-2,1,-1,0,0,1,0,1,0,0,0,1,1,0,0,1,0,1])
print "RREF(A)="
print A.rref()
print A.right_kernel()
print "nullity(A)=",; print A.right_nullity()
Output
RREF(A)=
[ 1 -3 0 0 0]
[ 0 0 1 0 0]
[ 0 0 0 1 0]
[ 0 0 0 0 1]
[ 0 0 0 0 0]
[ 0 0 0 0 0]
Free module of degree 5 and rank 1 over Integer Ring
Echelon basis matrix:
[3 1 0 0 0]
nullity(A)= 1 :OK ■
2015.10.29. Solved by 주영은 2015.10.29. Revised by 서승완 2015.10.29. Finalized by 주민규 2015.10.31.
Refinalized by 박시현 2015.11.4. Refinalized by 김원경 2015.11.8. Refinalized by 이승열 Final Ok by SGLee
Chapter7 Problem6
(old) For the following matrix , find a basis for its column space and compute the column rank .
Sol)
By using a Sage Find .
∴ ■
Double checked by Sage> http://math3.skku.ac.kr/home/pub/71 published by 주영은
http://math3.skku.ac.kr/home/pub/76/ revised by 박시현
Input |
A = matrix(6,5,[1,0,2,0,0,0,1,-1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,]) print A.rref() print A.rank() print A.right_nullity() |
Output |
[ 1 0 2 0 0] [ 0 1 -1 0 0] [ 0 0 0 1 0] [ 0 0 0 0 1] [ 0 0 0 0 0] [ 0 0 0 0 0] 4 1 ■ |
Comment: 의 RREF를 통해 행공간의 기저 역시 4차원임을 알 수 있었다. 또한, 의 rank를 직접 구해본 결과 같은 차원임을 확인 할 수 있었다.
마지막으로 의 해공간의 차원이 1이므로 Rank-Nullity 정리 가 성립함을 확인 할 수 있었다. ■
2015.10.22. Solved by 강민혁 2015.10.31. Revised by 김도희 & Finalized by 이종현 2015.11.9.
Refinalized by WK.Kim 2015.11.10. Refinalized by SC.Kim & Refinalized by 이승열 2015.11.15. Refinalized by WK.Kim Final OK by SGLee
(original) For given matrix compute its rank and nullity. Verify if the rank and nullity of satisfy the Rank-Nullity Theorem.
(1) (2)
Solution.
(1) =RREF()=
as , , are free variables
===++
∴ rank()=3, nullity()=3
∴ dim()=6=rank()+nullity(). ■
(2) =RREF()=
as is a free variable
===
∴ rank(A)=3, nulity(A)=1
∴ dim()=4=rank()+nulity(). ■
Double checked by Sage)
http://math3.skku.ac.kr/home/pub/102 by WK.Kim
A=matrix([[2, 5, 7, 9, 10, 11], [2, 3, 1, 2, 4, 8], [8, 6, 2, 1, 2, 3]])
print "RREF(A)="
print A.rref()
print "rank(A)="
print A.rank()
print "nullity(A)="
print A.right_nullity()
RREF(A)=
[ 1 0 0 -3/4 -3/2 -13/4]
[ 0 1 0 7/8 9/4 21/4]
[ 0 0 1 7/8 1/4 -5/4]
rank(A)=
3 ok.
nullity(A)=
3 ok. ■
A=matrix([[1, -1, 2, 1], [-1, 0, -1, 2], [2, -4, 6, 0], [3, 3, 0, -1], [0, -1, 1, 1]])
print "RREF(A)="
print A.rref()
print "rank(A)="
print A.rank()
print "nullity(A)="
print A.right_nullity()
RREF(A)=
[ 1 0 1 0]
[ 0 1 -1 0]
[ 0 0 0 1]
[ 0 0 0 0]
[ 0 0 0 0]
rank(A)=
3 ok.
nullity(A)=
1 ok. ■
2015.11.08. Solved 이승열 2015.11.08. Revised by SC.Kim
2015.11.09. Finalized by WK.Kim 2015.11.15. Refinalized by WK.Kim Final OK by SGLee
(New) For given matrix compute its rank and nullity.
Verify if the rank and nullity of satisfy the Rank-Nullity Theorem.
(1) (2)
Sol) (1)
(By. Sage http://math3.skku.ac.kr/home/pub/93 by SC.Kim)
⇒ . Number of free variables . . ■
(2) (By. Sage http://math3.skku.ac.kr/home/pub/94 by SC.Kim)
⇒ . Number of free variables . . ■
Double checked by Sage
(1) http://math3.skku.ac.kr/home/pub/91 by 이승열
A=matrix(QQ,3,6,[2,1,0,5,3,7,5,6,7,1,2,0,9,2,8,6,2,3])
print A.rank()
print A.right_nullity()
A.rank()+A.right_nullity()==6#number of column is 6
Answer:
3
3
True ok. ■
(2) http://math3.skku.ac.kr/home/pub/92 by 이승열
A=matrix(QQ,5,3,[1,2,6,10,2,7,9,8,7,9,8,3,1,2,5])
print A.rank()
print A.right_nullity()
A.rank()+A.right_nullity()==3#number of column is 3
Answer:
3
0
True ok. ■
2015.10.30. Solved by 박시현 2015.10.30. Revised by 남기현 2015.11.4. Finalized by WK.Kim 2015.11.09.
Refinalized by SC.Kim 2015.11.10..Refinalzied by 이승열 Final OK by SGLee
Check if .
Sol)
■
Double check by Sage)
http://math3.skku.ac.kr/home/pub/78/ by 이반석
A=matrix(QQ,4,5,[1,-2,1,1,2,0,1,1,3,4,1,2,5,13,5,-1,3,0,2,-2])
B=A.transpose()
print A.echelon_form()
print B.echelon_form()
print A.rank()
print B.rank()
print A.rank()==B.rank()
[1 0 3 7 0]
[0 1 1 3 0]
[0 0 0 0 1]
[0 0 0 0 0]
[ 1 0 0 -17/13]
[ 0 1 0 -3/13]
[ 0 0 1 4/13]
[ 0 0 0 0]
[ 0 0 0 0]
3
3
True ok ■
2015.11.16. Solved by SC.Kim 2015.11.16. Revised by 남기현 2015.11.16. Finalized by 서승완
2015.11.20. Refinalzied by 이승열 Final OK by SGLee
(New) Using the table below compute the rank of for matrix :
|
(a) |
(b) |
(c) |
(d) |
(e) |
size of |
|
|
|
|
|
|
3 |
2 |
1 |
3 |
2 |
Solution)
Let .
(i) rank of
(ii) rank of
(iii) rank of
(iv) rank of
Ans)
(a)
(i)
(ii)
(iii)
(iv) .
(c)
(i)
(ii)
(iii)
(iv) .
(e)
(i)
(ii)
(iii)
(iv) .
(b)
(i)
(ii)
(iii)
(iv) .
(d)
(i)
(ii)
(iii)
(iv) .
OK ■
2015.11.9. Solved by 서승완 2015.11.10. Finalize by 주민규 Final OK by SGLee
LA Chapter 7 Exercises 10
For , if , we say that has full row ran, and if , is said to have full column rank.
Determine if has full row rank and/or full column rank:
.
Solution)
. 4, .
: full column rank ■
Sage) http://math3.skku.ac.kr/home/pub/103/ by 서승완
: OK
■
2015.11.08. Solved by SC.Kim 2015.11.10. Finalized by WK.Kim 2015.11.23. Refinalzied by 이승열 Final Ok by SGLee
(NEW) For , find a basis and the dimension of the hyperplane .
Sol) For any in , .
=> => (since ).
A set {(3, 0, -1, 0), (0, 2, 0 ,-1), (1, 0, 1, -1)} forms a basis for hyperplane .■.
2015.10.30 solved by 강민혁 2015.10.31 revised by 정수호 2015.10.31 revised by 이반석 2015.11.4. Finalized by 이승열 2015.11.16. Refinalized by SC.Kim
Final OK by SGLee
(Original) For and , find the standard matrix for .
Sol)
, and
⇒
⇒
Ans)
■
2015.11.15. Solved by 이승열 2015.11.23. Finalized by 이승열 Final Ok by SGLee
For and , using , find and such
that , and .
Sol)
■
2015.10.30. Solved by 강민혁 2015.10.30. Revised by 김준현 2015.10.31. Revised by 이반석 2015.11.01 Finalized by 박원기 2015.11.04.
Refinalized by 이승열 2015.11.16. Refinalized by SC.Kim Final OK by SGLee
For given express as for which is in the direction of and is perpendicular to .
Sol)
. ■
2015.11.16. Solved by SC.Kim 2015.11.16. Revised by 남기현 2015.11.16. Finalized by 서승완 2015.11.16. Finalized by SC.Kim 2015.11.20. Refinalzied by 이승열 Final OK by SGLee
Chapter 7. Problem 15
For the following and , find the least squares solution to :
(1) , .
(2) , .
Solution)
Let . If , then .
The is the least squares solution.
(1) Solved by Sage.
http://math3.skku.ac.kr/home/pub/112
by SC.Kim
Input)
Ans) . OK ■
(2) Solved by Sage.
http://math3.skku.ac.kr/home/pub/113
by SC.Kim
Input)
Ans) There is no . OK ■
2015.11.16. Solved by SC.Kim 2015.11.16. Finalized by 남기현 2015.11.23. Refinlazied by 이승열 Final Ok by SGLee
Find the least squares curve passing through the five points .
Sol)
Let .
In the same way,
OK
Solve by Sage. http://math3.skku.ac.kr/home/pub/114 by SC.Kim
Input)
A=matrix(5, 5, [1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 1, 3, 9, 27, 81, 1, 4, 16, 64, 256, 1, 5, 25, 125, 625])
b=matrix(5, 1, [5, 1, -3, 1, 2])
print "A.inverse()*b="
print A.inverse()*b
Output)
A.inverse()*b=
[ -18]
[ 201/4]
[-857/24]
[ 37/4]
[ -19/24]
Ans) . OK ■
Note: 만일 3차식 을 구하라고 한다면 8장에서 배운 generalized inverse를 이용하면 구할 수 있다.
2015.10.26. Solved by 김준현 2015.10.27. Revised by SGLee 2015.10.29. Finalized by 김준현 2015.11.09. Refinalized by WK.Kim
2015.11.10. Refinalzied by SC.Kim 2015.11.10. Refinalzied by 이승열 Final OK by SGLee
Determine the values of which make the set orthogonal.
Sol)
⇒
⇒
, ■
2015.11.08. Solved by 이승열 2015.11.09. Finalized by WK.Kim 2015.11.16. Refinalized by SC.Kim Final OK by SGLee
(New) Find the orthonormal set relative to the following orthogonal vectors:
.
Sol)
Orthonormal set .■
Double checked by Sage) http://math3.skku.ac.kr/home/pub/95 by SC.Kim
v1=vector([1,0,1])
v2=vector([0,1,0])
v3=vector([-1,0,1])
llv1ll=v1.norm()
llv2ll=v2.norm()
llv3ll=v3.norm()
print "llv1ll=",; print llv1ll
print "llv2ll=",; print llv2ll
print "llv3ll=",; print llv3ll
print "x1=",; print v1/llv1ll
print "x2=",; print v2/llv2ll
print "x3=",; print v3/llv3ll
Ans)
llv1ll= sqrt(2)
llv2ll= 1
llv3ll= sqrt(2)
x1= (1/2*sqrt(2), 0, 1/2*sqrt(2))
x2= (0, 1, 0)
x3= (-1/2*sqrt(2), 0, 1/2*sqrt(2)) : OK. ■
2015.10.27. Solved by 김준현 2015.11.15. Refinalized by WK.Kim 2015.10.27. Revised by SGLee 2015.11.20. Refinalized by WK.Kim
2015.10.29. Finalized by 김준현 2015.11.23. Refinalzied by 이승열 Final OK by SGLee
Show that each of the following sets of vectors is linearly independent, and find its corresponding orthonormal set:
(1) .
(2) .
Sol)
(1)
let
⇒
⇒
⇒ ∴linearly independent. OK
Using Gram-Schmidt orthonormal process, K
⇒
⇒
⇒
⇒
∴ OK ■
(2)
let
⇒
⇒
⇒
⇒ ∴linearly independent.
Using Gram-Schmidt orthonormal process, OK
⇒
⇒
⇒
⇒
OK ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/131 by WK.Kim
1) |
2) |
x1=vector([0,0,1,0]) x2=vector([1,0,1,1]) x3=vector([1,1,2,1]) A=matrix([x1,x2,x3]) [G,mu]=A.gram_schmidt() B=matrix([G.row(i)/G.row(i).norm() for i in range(0,3)]);B |
x1=vector([1,1,1,1]) x2=vector([-1,4,4,-1]) x3=vector([4,-2,2,0]) A=matrix([x1,x2,x3]) [G,mu]=A.gram_schmidt() B=matrix([G.row(i)/G.row(i).norm() for i in range(0,3)]);B |
[ 0 0 1 0] [1/2*sqrt(2) 0 0 1/2*sqrt(2)] [ 0 1 0 0] : ok |
[ 1/2 1/2 1/2 1/2] [-1/2 1/2 1/2 -1/2] [ 1/2 -1/2 1/2 –1/2] : ok |
2015.10.16. Solved by 강민혁 2015.11.04. Revised by 정범진 2015.11.09. finalized by 김선호 2015.11.10. Refinalized by WK.Kim 2015.11.15. Refinalized by 이승열
For given plane : {} and vector v=(2,4,-3), find the following (Note that the inner product is defined to be .
(1) A basis for the 2-dimensional vector space represented by plane and its corresponding orthonormal basis
(2) projv
Sol)
(1)
a basis: (a basis for )
Using Gram-Schmidt
⇒ ,
⇒
■
(2)
■
2015.10.26. Solved by 강민혁 2015.11.04. Revised by 정범진 2015.11.06. FInalized by 정양헌 2015.11.09. Refinalized by 정양헌 Final by TA Lee
For the ordered basis ={(0,1,1,1),(1,0,-1,1),(1,2,0,2),(3,-2,2,0)} for .
(1) For x=(7,-7,5,4), find its coordinate vector [x] relative to .
(2) For y=(1,-4,4,3), find its coordinate vector [y] relative to .
(3) Find the coordinate vector [2x+y] of 2x+y relative to .
(4) For the above x and y, ind [3x] and [-5y].
Sol)
(1),(2)
(3)
(4) ■
2015.10.26. Solved by 강민혁 2015.11.04. Revised by 정범진 2015.11.06. Finalized by 정양헌
2015.11.09. Refinalized by 정양헌 2015.11.10. Refinalized by WK.Kim 2015.11.16. Refinalized by SC.Kim Final OK by SGLee
For u=(1,2), u=(2,3), v=(1,3), v=(1,4), let ={u,u}, ={v,v} which are bases for .
(1) Find the transition matrix .
(2) Find the transition matrix .
(3) Suppose [w]= (1,1). Find [w] using the transition matrix .
(4) Suppose [w]= (3,2). Find [w] using the transition matrix .
Sol)
(1)
⇒
RREF form
⇒
■
(2)
■
(3)
■
(4)
■
2015.11.08. Solved by SC.Kim 2015.11.08. Revised by 이승열 2015.11.09. Finalized by WK.Kim
2015.11.15. Refinalized by WK.Kim Final OK by SGLee
If the size of matrix is , what is the value of
?
Sol)
Ans:
■
2015.11.08. Solved by SC.Kim 2015.11.08. Revised by H.Haziq 2015.11.10. Finalized by 이승열
2015.11.15. Refinalized by WK.Kim Final OK by SGLee
(Select the right one statement) If one replaces a matrix with its transpose, then
A. The image may change, but the kernel, rank, and nullity do not change.
B. The image, kernel, rank, and nullity may all change.
C. The image, rank, and kernel may change, but the nullity does not change.
D. The image, kernel, rank, and nullity all do not change.
E. The image, kernel, and nullity may change, but the rank does not change.
F. The kernel may change, but the image, rank, and nullity do not change.
G. The image and kernel may change, but the rank and nullity do not change.
Sol) Let . Then
(1) (Theorem 7.22)
(2)
With (1) and (2)
Ans) E. The image, kernel, and nullity may change, but the rank does not change.” ■
2015.11.08. Solved by SC.Kim 2015.11.10. Revised by H.Haziq 2015.11.11. Finalized by 이승열
2015.11.18. Refinalized by WK.Kim Final OK by SGLee
(Select one) Let be a linear transformation. Then
A. is invertible if and only if the rank is five.
B. is one-to-one if and only if the rank is three; is never onto.
C. is onto if and only if the rank is two; is never one-to-one.
D. is one-to-one if and only if the rank is two; is never onto.
E. is onto if and only if the rank is three; is never one-to-one.
F. is onto if and only if the rank is five; is never one-to-one.
G. is one-to-one if and only if the rank is five; is never onto.
Sol)
1)
For a linear transformation ,
if is one-to-one and onto, then . (∵ By the Definition of Isomorphism)
∴ ⇒ is never onto.
2)
The defined by a matrix and is one-to-one
⇒
∴
Ans) B. is one-to-one if and only if the rank is three; is never onto. OK ■
2015.11.23. Solved by 이승열 2015.11.23. Finalized by 이승열
(Select one) If a linear transformation is onto, then
A. The rank is three and the nullity is zero.
B. The rank and nullity can be any pair of non-negative numbers that add up to three.
C. The rank is three and the nullity is two.
D. The rank is two and the nullity is three.
E. The situation is impossible.
F. The rank and nullity can be any pair of non-negative numbers that add up to five.
G. The rank is five and the nullity is two.
Ans: E
Note:
is onto.
(contradiction)
The situation is impossible.
2015.11.08. Solved by SC.Kim 2015.11.10. Revised by H.Haziq 2015.11.10. Refinalized by 이승열
2015.11.15. Refinalized by WK.Kim Final OK by SGLee
(Select one) If a linear transformation is one-to-one, then
A. The rank is five and the nullity is two.
B. The situation is impossible.
C. The rank and nullity can be any pair of non-negative numbers that add up to
five.
D. The rank is two and the nullity is three.
E. The rank is three and the nullity is zero.
F. The rank is three and the nullity is two.
G. The rank and nullity can be any pair of non-negative numbers that add up to
three.
Sol)
The defined by a matrix
If is one-to-one, then
⇒
Ans) F. The rank is three and the nullity is two. ■
2015.11.16. solved by WK.Kim 2015.11.16. revised by 이반석 2015.11.19. Finalized by 주민규
2015.11.23. Refinalzied by 이승열 Final OK by SGLee
(1) If the homogeneous linear system has linear equations and unknowns, what is the maximum possible value for the dimension of the solution space?
Sol)
⇒ minimum value of .
∴ maximum value of Ok ■
(2) What is the dimension of a hyperplane in perpendicular to a vector in ?
Sol)
, where Ok ■
(3) List all of the possible dimensions of subspaces of .
Ans) 0, 1, 2, 3, 4, 5, 6 Ok ■
(4) What is the dimension of the subspace of spanned by the three vectors , , ?
Sol)
⇒
(dimension of the subspace) Ok ■
2015.11.15. Solved by 이승열 2015.11.23. Finalized by 이승열 Final Ok by SGLee
Suppose is a basis for . If is an invertible matrix of order ,
show that the set is also a basis for .
Sol)
( is linear independent.)
is a linear independent set of n vectors in
is a basis for . ■
2015.11.08. Solved by SC.Kim 2015.11.10. Revised by H.Haziq
2015.11.15. Finalized by 이승열 Final OK by SGLee
Determine if the following matrix and have the same null space and row space:
Sol) Solve by Sage http://math3.skku.ac.kr/home/pub/96 by SC.Kim
Input)
A=matrix(3, 2, [1, 2, 2, 4, -1, -2])
B= A.transpose()
C= B*A
print "A="
print A
print "B"
print B
print "BA=C="
print C
print "row space of A"
print A.row_space()
print A.right_nullity()
print "row space of C"
print C.row_space()
print C.right_nullity()
Output)
A=
[ 1 2]
[ 2 4]
[-1 -2]
B
[ 1 2 -1]
[ 2 4 -2]
BA=C=
[ 6 12]
[12 24]
row space of A
Free module of degree 2 and rank 1 over Integer Ring
Echelon basis matrix:
[1 2]
1
row space of C
Free module of degree 2 and rank 1 over Integer Ring
Echelon basis matrix:
[ 6 12]
1
Ans) ,
but row space of and are different. ■
2015.11.09. Solved by SC.Kim 2015.11.10. Finalize by MK.Ju Final OK by SGLee
What happens if the Gram-Schmidt Orthonormalization Procedure is applied to linearly dependent vectors?
Sol)
One base is removed. So we can’t get orthonormal basis. ■
Checked by Sage)
(http://math3.skku.ac.kr/home/pub/105 by SC.Kim, MK.Ju)
Input
x1=vector([1,1,0])
x2=vector([0,1,2])
x3=vector([1,2,2])
A=matrix([x1,x2,x3])
[G,mu]=A.gram_schmidt()
B=matrix([G.row(i) / G.row(i).norm() for i in range(0, 2)]); B
P1= B*B.transpose()
P2= B.transpose()*B
print G
print P1
print P2
print P1==P2
Output
[ 1 1 0]
[-1/2 1/2 2]
[1 0]
[0 1]
[ 5/9 4/9 -2/9]
[ 4/9 5/9 2/9]
[-2/9 2/9 8/9]
False
, but ■
2015.11.23. Solved by 이승열 2015.11.23. Finalzied by 이승열 Final OK by SGLee
Suppose the columns of are linearly independent.
What you can tell about the column spaces of and .
Sol)
of columns
Columns vectors in in
Columns vectors in in
Columns of forms a basis in &
Columns of forms a dimensional column space in . ■
2015.11.16. Solved by SC.Kim 2015.11.16. Finalized by 남기현 2015.11.16. Refinalized by 서승완 2015.11.16.
Refinalized by WK.Kim 2015.11.16. Refinalized by SC.Kim 2015.11.20. Refinalized by 이승열 Final Ok by SGLee
Show that the set spans .
Solution)
Let . Then, .
. Therefore are linearly independent.
By Theorem 7.4.9 (8) and (9), spans .
Ok ■
Double checked by Sage)
http://math3.skku.ac.kr/home/pub/125 by SC.Kim
: OK
■
2015.11.16. solved by WK.Kim 2015.11.23. Finalzied by 이승열 Final OK by SGLee
(New)What are the all possible ranks of according to the varying values of ?
.
Sol)
■
2015. 11. 09 Solved by Lee Naeul, 2015. 11. 11 Finalized by Jung Yanghun 2015.11.11. Refinalized by MK .Ju, Final OK by SGLee
Suppose a linear transformation is defined by and is an ordered basis for .
Find the matrix representation of relative to the ordered basis .
Sol)
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/98/ by Lee Naeul
x,y = var('x, y') h(x,y) = [x-y, x+y] T = linear_transformation(QQ^2, QQ^2, h) x1 = vector([1,1]) x2 = vector([-1,0]) y1 = vector([1,1]) y2 = vector([-1,0]) B = column_matrix([y1, y2, T(x1), T(x2)]) print B C = B.echelon_form() print C A = C.submatrix(0,2,2,2) print A |
[ 1 -1 0 -1] [ 1 0 2 -1]
[ 1 0 2 -1] [ 0 1 2 0]
[ 2 -1] [ 2 0]
|
:OK ■
2015.11.14. Solved by 변희성 2015.11.14. Revised by 서승완 2015.11.14. Finalized by 주민규 Final OK by SGLee
Chapter 8 Problem 1
Suppose a linear transformation is defined by and is an ordered basis for .
Find the matrix representation of relative to the ordered basis .
Solution)
Let
■
Sage) http://math3.skku.ac.kr/home/math2013/102/ by 서승완
:OK ■
2015.11.20. Solved by 서승완 2015.11.20. Revised by 남기현 2015.11.21.Finalized by 백지원 Final OK by SGLee
Suppose a linear transformation is defined by and is an ordered basis for .
Find the matrix representation of relative to the ordered basis .
Solution)
,
RREF by Sage:
■
Sage) http://math3.skku.ac.kr/home/pub/130/ by 서승완
2015. 11. 09 Solved by Lee Naeul 2015. 11. 11 Finalized by Jung Yanghun 2015.11.11. Refinalized by MK .Ju Final OK by SGLee
Suppose a linear transformation is defined by and is an ordered basis for .
Find the matrix representation of relative to the ordered basis .
Sol)
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/98/ by Lee Naeul
x,y = var('x, y') h(x,y) = [x-y, x+y] T = linear_transformation(QQ^2, QQ^2, h) x1 = vector([1,1]) x2 = vector([-1,0]) y1 = vector([1,1]) y2 = vector([-1,0]) B = column_matrix([y1, y2, T(x1), T(x2)]) print B C = B.echelon_form() print C A = C.submatrix(0,2,2,2) print A |
[ 1 -1 0 -1] [ 1 0 2 -1]
[ 1 0 2 -1] [ 0 1 2 0]
[ 2 -1] [ 2 0]
|
:OK ■
2015.11.14. solved by 변희성, 2015.11.14. Revised by 서승완, 2015.11.15. Finalized by 주민규, 2015.12.10. Refinalized by 백지원+강민혁 Final OK by SGLee
Let be defined by and let , be ordered bases for , respectively, where
, , , , .
Find the matrix representation of with respect to the ordered bases and
Solution)
.
= .
■
Sage) http://math3.skku.ac.kr/home/pub/122/ by 서승완
x, y = var('x, y') h(x, y) = [x-2*y,5*x-y,2*x+3*y] T = linear_transformation(QQ^2, QQ^3, h) x1=vector([1, 0]) x2=vector([0, 1]) y1=vector([0,0,1]) y2=vector([0,1,0]) y3=vector([1,0,0]) B=column_matrix([y1, y2, y3, T(x1), T(x2)]) print B C=B.echelon_form() print C A=C.submatrix(0, 3, 3, 2) print A |
[0 0 1 1 -2] [0 1 0 5 -1] [1 0 0 2 3]
[1 0 0 2 3] [0 1 0 5 -1] [0 0 1 1 -2]
[2 3] [5 -1] [1 -2] : OK |
2015.11.20. Solved by 서승완 2015.11.20. Revised by 남기현 2015.11.20. Finalized by 변희성 Final OK by TA. Lee
LA Chapter 8 Exercises 2(new)
Let be defined by and let , be ordered bases for , respectively, where
, , , , , .
Find the matrix representation of with respect to the ordered bases and .
Solution)
Since ,, we first compute
.
Then we have
We can get its RREF by Sage:
■
Sage) http://math3.skku.ac.kr/home/pub/132/ by 서승완
x, y, z = var('x, y, z') h(x, y, z) = [3*x-2*z,6*y-2*z] T = linear_transformation(QQ^3, QQ^2, h) x1=vector([0,0,1]) x2=vector([0,1,1]) x3=vector([-1,0,-2]) y1=vector([2,0]) y2=vector([0,1]) B=column_matrix([y1, y2, T(x1), T(x2), T(x3)]) print B C=B.echelon_form() print C A=C.submatrix(0, 2, 2, 3) print A |
[ 2 0 -2 -2 1] [ 0 1 -2 4 4]
[ 1 0 -1 -1 1/2] [ 0 1 -2 4 4]
[ -1 -1 1/2] [ -2 4 4] |
2015.11.14. solved by 변희성 2015.11.14. Revised by 서승완 2015.11.14. Finalized by 주민규 Final OK by SGLee
Suppose a linear transformation is defined by and , are ordered bases for .
(1) Find the matrix representation of relative to the ordered basis .
(2) Find the transition matrix from to .
(3) Compute .
Solution)
(1)
■
(2)
■
(3)
■
2015.11.20. Solved by 서승완, 2015.11.20. Revised by 남기현 2015.11.20. Finalized by 변희성, Final OK by TA.Lee
Suppose a linear transformation is defined by
and ,
are ordered bases for
(1) Find the matrix representation of relative to the ordered basis .
(2) Find the transition matrix from to .
(3) Compute .
Solution) (1)
Therefore we get
we can have its RREF by Sage
■
(2)
,
■
(3) ■
Sage) http://math3.skku.ac.kr/home/pub/133/ by 서승완
x, y, z = var('x, y, z') h(x, y, z) = [-x-6*z,3*x+4*z,-2*x+5*y] T = linear_transformation(QQ^3, QQ^3, h) x1=vector([1,0,0]) x2=vector([0,0,1]) x3=vector([1,1,0]) y1=vector([2,0]) y2=vector([0,1]) B=column_matrix([x1, x2, x3, T(x1), T(x2), T(x3)]) print B C=B.echelon_form() print C A=C.submatrix(0, 3, 3, 3) print A P=matrix(3,3,[0,-3,3,-1,3,2,0,1,-3]) print P P1=P.inverse() print P1 print P1*A*P |
[ 1 0 1 -1 -6 -1] [ 0 0 1 3 4 3] [ 0 1 0 -2 0 3]
[ 1 0 0 -4 -10 -4] [ 0 1 0 -2 0 3] [ 0 0 1 3 4 3]
[ -4 -10 -4] [ -2 0 3] [ 3 4 3]
[ 0 -3 3] [-1 3 2] [ 0 1 -3]
[-11/6 -1 -5/2] [ -1/2 0 -1/2] [ -1/6 0 -1/2]
[-25/3 49/3 95/3] [ -3 8 6] [ 1/3 2/3 -2/3] |
2015.11.12. Solved by 김준현 2015.11.13. Revised by 이반석 2015.11.14. Finalized by 이나을 Final OK by SGLee
Determine if the given matrix is diagonalizable. If is diagonalizable, find matrix diagonalizing and the associated diagonal matrix such that .
(1) .
(2) .
Sol)
(1)
eigenvector of :
eigenvector of :
matrix has 2 linear independent eigenvectors.
∴ A is diagonalizable.
(2)
eigenvector of :
eigenvector of : ,
matrix has 3 linear independent eigenvectors.
∴ is diagonalizable.
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/117/ by Lee Naeul
A = matrix([[3,-4],[-4,3]]) print A print A.eigenvalues() E = identity_matrix(2) print (A-7*E).echelon_form() print (A-(-1)*E).echelon_form() x1 = vector([1,-1]) x2 = vector([1,1]) P = column_matrix([x1, x2]) print P print P^-1*A*P |
[ 3 -4] [-4 3]
[7, -1]
[4 4] [0 0]
[ 4 -4] [ 0 0]
[ 1 1] [-1 1]
[ 7 0] [ 0 -1] : OK |
B = matrix([[2,1,1],[1,2,1], [1,1,2]]) print B print B.eigenvalues() E = identity_matrix(3) print (B-4*E).echelon_form() print (B-E).echelon_form() print (B-E).echelon_form() x1 = vector([1,1,1]) x2 = vector([-1,1,0]) x3 = vector([-1,0,1]) P = column_matrix([x1, x2, x3]) print P print P^-1*B*P |
[2 1 1] [1 2 1] [1 1 2]
[4, 1, 1]
[ 1 1 -2] [ 0 3 -3] [ 0 0 0]
[1 1 1] [0 0 0] [0 0 0]
[1 1 1] [0 0 0] [0 0 0]
[ 1 -1 -1] [ 1 1 0] [ 1 0 1]
[4 0 0] [0 1 0] [0 0 1] : OK |
■
2015. 11. 09 Solved by Lee Naeul 2015. 11. 14 Revised by 이반석 2015.11.14. Finalized by 주민규 Final OK by SGLee
Find the algebraic and geometric multiplicity of each eigenvalue of :
.
Sol)
Characteristic equation of
Thus, eigenvalues are 1 and 2.
'1' has algebraic multiplicity 2 and geometric multiplicity 1.
'2' has algebraic multiplicity 1.
■ Double Checked by Sage) http://math3.skku.ac.kr/home/pub/118/ by Lee Naeul
http://math3.skku.ac.kr/home/pub/120/ by 이반석
A=matrix(QQ,3,3,[1,0,-1,0,1,2,2,1,2]) print "λ= "+str(A.eigenvalues()) ev=A.eigenvalues() temp=0 i=0 for sol in ev: if(sol != temp): i=ev.count(sol) print str(sol) + " have " + str(i) + " algebraic multiplicity." temp=sol A.eigenvectors_right() |
λ= [2, 1, 1] 2 have 1 algebraic multiplicity. 1 have 2 algebraic multiplicity. [(2, [(1, -2, -1)], 1), (1, [(1, -2, 0)], 2)] |
: OK ■
2015.11.09. Solved by 김준현 2015.11.10. Revised by 주민규 2015.11.10. Finalized by 박원기+백지원 Final OK by SGLee
Find the algebraic and geometric multiplicity of each eigenvalue of :
.
Solution)
Characteristic polynomial of
i)
∴L.I. eigenvector:
ii)
∴L.I. eigenvector:
∴ eigenvalue of : 1, 2
each of algebraic multiplicity: 2, 1 respectively
each of geometric multiplicity: 1, 1 respectively ■
2015.11.29. solved by 백지원 2015.11.29. revised by 이나을 2015.11.30. finalized by 강민혁 Final OK by SGLee
Find matrix orthogonally diagonalizing matrix and the diagonal matrix such that , using Sage.
.
Sol)
When .
When : two o.g. (linear independent) vectors
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/148/ by Leenaeul
A=matrix(QQ, [[1,2,2],[2,1,-2],[2,-2,1]]) print A.eigenvalues() print A.eigenvectors_right() |
[-3, 3, 3]
[(-3, [(1, -1, -1)], 1), (3, [(1, 0, 1), (0, 1, -1)], 2)] |
2015.11.29. solved by 이나을, 2015.11.29. revised by 백지원 2015.11.30. finalized by 강민혁, 2015.12.07 Refinalized by 이나을 Final OK by SGLee
In each of the following matrix and set of linear independent eigenvectors of are given. Find an orthogonal matrix and a diagonal matrix such that .
(1) , .
(2) , .
Sol)
(1) Let
Since and are orthogonal, we find and by normalizing them.
,
(2) Let
Since and are orthogonal, we find and by normalizing them.
, ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/149/ by Lee Naeul
1) A=matrix(QQ, 2,2, [-3,6,6,-3]) x1 = vector([-2, 2]) z1 = x1 / x1.norm()
x2 = vector([5,5]) z2 = x2 / x2.norm()
P = column_matrix([z1, z2]) print "Matrix P : " print P print "Matrix D : " print P.transpose() * A * P |
Matrix P : [-1/2*sqrt(2) 1/2*sqrt(2)] [ 1/2*sqrt(2) 1/2*sqrt(2)]
Matrix D : [-9 0] [ 0 3] |
2) A=matrix(QQ, 2,2, [7,2,2,4]) x1 = vector([2, 1]) z1 = x1 / x1.norm() x2 = vector([1,-2]) z2 = x2 / x2.norm()
P = column_matrix([z1, z2]) print "Matrix P : " print P print "Matrix D : " print P.transpose() * A * P |
Matrix P : [ 2/5*sqrt(5) 1/5*sqrt(5)] [ 1/5*sqrt(5) -2/5*sqrt(5)] Matrix D : [8 0] [0 3] |
■
2015.11.14 Solved by 정범진 2015.11.14 Revised by 이반석 2015.11.21. Finalized by 백지원
Compute , when
Sol)
■
2015.11.14. Solved by 정범진 2015.11.14. Revised by 이반석 2015.11.14. Finalized by 주민규+백지원 Final Okay by SGLee
Write the following expression as a quadratic form :
Sol)
,
,
■
2015. 11. 19 Solved by Lee Naeul, 2015. 11. 20 Revised by H.Haziq 2015.12.10. Refinalized by 강민혁, Final OK by SGLee
Eliminate the cross-product term from the following:
.
Sol) Convert the Quadratic equation into a matrix form.
=>
Let
when , .
when , .
Let
Let
OK █
2015. 11. 19 Solved by Lee Naeul 2015. 11. 20. Revised by H.Haziq Final OK by SGLee
Sketch the graph of the following equation:
Sol)
Let , ,
Solve the equation below,
,
,
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/127/ by Lee Naeul
var('x y') f=x^2+4*x*y+4*y^2+6*x+2*y-25 implicit_plot(f==0, (x,-10,10), (y,-10,10)) |
■
2015.11.29. solved by 백지원, 2015.11.29. revised by 이나을 2015.11.30. finalized by 강민혁
Eliminate the cross-product terms from the quadratic surface by properly rotating the axes.
Solution)
Let ,
Let
Then .
Eigenvectors are
. ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/153/ by Leenaeul
A=matrix(3,3,[5,2,0,2,6,2,0,2,7]) print "eigenvalues : " print A.eigenvalues() print "eigenvectors : " print A.eigenvectors_right() P = matrix(3,3,[1/3, 2/3, 2/3,2/3,1/3,-2/3,2/3,-2/3,1/3]) print "Matrix P : " print P print "PT*A*P : " print P.transpose()*A*P |
eigenvalues : [9, 6, 3]
eigenvectors : [(9, [(1, 2, 2)], 1), (6, [(1, 1/2, -1)], 1), (3, [(1, -1, 1/2)], 1)]
Matrix P : [ 1/3 2/3 2/3] [ 2/3 1/3 -2/3] [ 2/3 -2/3 1/3]
PT*A*P : [9 0 0] [0 6 0] [0 0 3] : OK |
2015. 11. 19 Solved by Lee Naeul 2015. 11. 20 Revised by H.Haziq Final OK by SGLee
Compute the singular values of matrix :
.
Sol)
A sigular value of matrix is . ■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/128/ by Lee Naeul
A=matrix([1, 2, 0]) B=A.transpose()*A eig=B.eigenvalues() sv=[sqrt(i) for i in eig] print sv |
[sqrt(5)] |
: OK ■
2015. 11. 19 Solved by Lee Naeul 2015. 11. 20 Revised by H.Haziq Final OK by SGLee
Find the SVD of :
.
Sol) i)
,
A unit eigenvector of correspoding to is .
A unit eigenvector of correspoding to is .
ii) . ,
Likewise, unit eigenvectors of are
, ,
, , ,
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/129/ by Lee Naeul
AA_t = matrix(QQ,2,2,[2,1,1,2]) A_tA = matrix(QQ,3,3,[1,1,0,1,2,1,0,1,1]) print AA_t.right_eigenvectors() print A_tA.right_eigenvectors() |
[(3, [(1, 1)], 1), (1, [(1, -1)], 1)] [(3, [(1, 2, 1)], 1), (1, [(1, 0, -1)], 1), (0, [(1, -1, 1)], 1)] |
2015.11.20. Solved by 정양헌, 2015.11.20. Finalized by 김준현 2015.11.23. Refinalized by 백지원, Final OK by SGLee
In the below the SVD of is given. Find .
.
Sol)
■
Checked by Sage) http://math1.skku.ac.kr/home/pub/2526/ by 김준현
A=matrix(2,2, [1/sqrt(2),-1/sqrt(2),1/sqrt(2),1/sqrt(2)])
B=matrix(3,2,[1/sqrt(3),0,0,1,0,0])
C=matrix(3,3, [1/sqrt(6),2/sqrt(6),1/sqrt(6),1/sqrt(2),0,-1/sqrt(2),1/sqrt(3),-1/sqrt(3),1/sqrt(3)])
D=matrix(2,3,[sqrt(3),0,0,0,1,0])
E=A.transpose()
F=C.transpose()
R= (F*B*E).apply_map(lambda x: x.simplify_full()) #pseudo-inverse
A= A*D*C #original matrix
print R
print (R*A*R).apply_map(lambda x: x.simplify_full())
[-1/3 2/3]
[ 1/3 1/3]
[ 2/3 -1/3]
[-1/3 2/3]
[ 1/3 1/3]
[ 2/3 –1/3] ■
2015.11.29. solved by 백지원 2015.11.29. revised by 이나을 2015.11.29. finalized by 강민혁 2015.12.07 Refinalized by 이나을 Final OK by SGLee
The following matrix has a full column rank. Find its pseudo-inverse.
.
Sol)
The matrix has a full column rank.
=> is an invertible matrix.
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/150/ by Lee Naeul
A = matrix(QQ, [[1,1], [0,2],[3,7]]) B = A.transpose() * A print B.inverse() * A.transpose() |
[ 4/7 -11/14 1/7] [ -3/14 5/14 1/14] : OK |
2015.11.20. Solved by 정양헌 2015.11.20. Revised by 남기현 2015.11.21. Finalized by 백지원 Final OK by SGLee
For given vectors , compute Euclidean inner products and .
Sol)
■
2015.11.20. Solved by 정양헌 2015.11.20. revised by 정다산 2015.11.21. finalized by 백지원 Final OK by SGLee
Let , be vectors in with Euclidean inner product defined. Compute the norms and , and .
Sol)
■
Double checked by Sage) http://math1.skku.ac.kr/home/pub/2525/
u=vector([62-85*I, 9, 66*I, 20*I+6, 63+11*I])
v=vector([31-55*I, 13-61*I, 21-63*I, 11*I, -71+9*I])
print u.norm()
print v.norm()
print (u-v).norm()
8*sqrt(313)
sqrt(17529)
sqrt(40757)
2015.11.29. solved by 백지원, 2015.11.29. revised by 이나을 2015.11.30. finalized by 강민혁, 2015.12.07 Refinalized by 이나을. Final OK by SGLee
New)
Find the eigenvalues of and a basis for the eigenspace associated with each eigenvalue.
Sol) i)
ii) When ,
: A basis for the eigenspace corresponding .
When
: A basis for the eigenspace corresponding . ■
( A basis : = {} corresponding eigenvalues = .)
A=matrix(QQ, 2,2, [-3,6,6,-3]) print "eigenvalues : " print A.eigenvalues() print "eigenvectors : " print A.eigenvectors_right() |
eigenvalues : [3, -9]
eigenvectors : [(3, [(1, 1)], 1), (-9, [(1, -1)], 1)] : OK |
2015.11.29. solved by 이나을 2015.11.29. revised by 백지원, 2015.11.30. finalized by 강민혁 2015.12.07 refinalized by 이나을, Final OK by SGLee
Find any invertible matrix diagonalizing a given matrix which has complex eigenvalues?
.
Solution)
When ,
=>
. Then by Theorem 8.7.1.
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/152/ by Leenaeul
A=matrix([[6,-4],[8,-2]]) print "eigenvalues of A : " print A.eigenvalues() print "eigenvectors of A : " print A.eigenvectors_right() |
eigenvalues of A : [2 - 4*I, 2 + 4*I]
eigenvectors of A : [(2 - 4*I, [(1, 1 + 1*I)], 1), (2 + 4*I, [(1, 1 - 1*I)], 1)] : OK |
2015.11.21. Solved by 정양헌 2015.11.21. Revised by 이반석 2015.11.21. Finalized by 백지원 2015.11.22. Refinalized by 이나을 Final OK by SGLee
Find the conjugate transpose of the following matrix :
.
Sol)
█
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/134/ by Lee Naeul
U=matrix(4,2,[3+i,-2*i,1,6*i,3*i,1+i,4,-1+5*i]) print U U.conjugate_transpose() |
[ I + 3 -2*I] [ 1 6*I] [ 3*I I + 1] [ 4 5*I - 1]
[ -I + 3 1 -3*I 4] [ 2*I -6*I -I + 1 -5*I - 1] |
█
2015.11.21. Solved by 정양헌 2015.11.21. Revised by 이반석 2015.11.21. Finallized by 정진규 2015.11.22. Refinalized by 이나을 2015.11.23. Refinalized by 정양헌 Final OK by TA Lee
Determine which matrices in the below are Hermitian.
(a) (b)
(c) (d)
(e) (f)
Sol) If a complex square matrix satisfying , is called a Hermitian matrix.
(a) (b)
(c)
(d) .
(e)
(f)
Answer: (b)
Note : If a complex square matrix satisfying , is called a Skew-Hermitian matrix. (e) is Skew-Hermitian matrix. █
2015.11.21. Solved by 정양헌, 2015.11.21. Revised by 이반석 2015.11.21.Finalized by 정진규, 2015.11.22.Refinalized by 이나을 2015.11.23. Refinalized by 정양헌, Final OK by SGLee
Determine if each matrix in the below is unitary.
(a) (b)
Sol)
If matrix is unitary, then ( is a nonsingular matrix)
(a)
(b)
Answer : (a) is unitary. █
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/137/ by Lee Naeul
A = matrix(2,2,[1/sqrt(2),1/sqrt(2),1/sqrt(2), -1/sqrt(2)]) print "Matrix A : " print A A_h = A.conjugate_transpose() print "Matrix A^* : " print A_h print "Matrix A*A^* : " print A*A_h |
Matrix A : [ 1/2*sqrt(2) 1/2*sqrt(2)] [ 1/2*sqrt(2) -1/2*sqrt(2)]
Matrix A^* : [ 1/2*sqrt(2) 1/2*sqrt(2)] [ 1/2*sqrt(2) -1/2*sqrt(2)]
Matrix A*A^* : [1 0] [0 1] : OK |
B = matrix(3,3,[0,i,1,i,0,2+i,-1,-2+i,0]) print "Matrix B : " print B B_h = B.conjugate_transpose() print "Matrix B^* : " print B_h print "Matrix B*B^* : " print B*B_h |
Matrix B : [ 0 I 1] [ I 0 I + 2] [ -1 I - 2 0]
Matrix B^* : [ 0 -I -1] [ -I 0 -I - 2] [ 1 -I + 2 0]
Matrix B*B^* : [ 2 -I + 2 -2*I + 1] [ I + 2 6 -I] [ 2*I + 1 I 6] : OK |
2015.11.21. Solved by 정양헌 2015.11.21. Revised by 이반석 2015.11.21. Revised by 정양헌 2015.11.21. Finalized by 백지원 2015.11.22. Refinalized by 이나을 2015.11.23. Refinalized by 정양헌 Final OK by TA Lee
Replace each by a complex number to make matrix Hermitian.
Sol)
If a complex square matrix satisfying , is called a Hermitian matrix
Let , we can denote unknown entries .
( is complex number.)
█
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/136/ by Lee Naeul
a, b, c = var('a b c')
A = matrix(3,3,[1,i,2-3*i,a,-3,1,b,c,2])
print "Matrix A : "
print A
A_h = A.conjugate_transpose()
print A_h
2015.11.21. Solved by 정양헌 2015.11.21. Revised by 이반석 2015.11.21.Finalized by 정진규 2015.11.22.Refinalized by 이나을 Final OK by TA Lee
Show that the following matrix is unitary, and find its inverse .
Sol)
If matrix is unitary, then
■
Double Checked by Sage) http://math3.skku.ac.kr/home/pub/135/ by Lee Naeul
A = matrix([[(sqrt(3)+i)/(2*sqrt(2)),(1-i*sqrt(3))/(2*sqrt(2))],[(1+i*sqrt(3))/(2*sqrt(2)),(i-sqrt(3))/(2*sqrt(2))]]) print "Matrix A : " print A B = A.conjugate_transpose() print "Matrix B : " print B |
Matrix A : [ 1/4*(sqrt(3) + I)*sqrt(2) 1/4*(-I*sqrt(3) + 1)*sqrt(2)] [ 1/4*(I*sqrt(3) + 1)*sqrt(2) -1/4*(sqrt(3) - I)*sqrt(2)]
Matrix B : [ 1/4*(sqrt(3) - I)*sqrt(2) 1/4*(-I*sqrt(3) + 1)*sqrt(2)] [ 1/4*(I*sqrt(3) + 1)*sqrt(2) -1/4*(sqrt(3) + I)*sqrt(2)] |
* Sage에서 행렬 연산은 유리화되어 표현되었습니다.
2015. 12. 10. Solved by 김이곤 2015. 12. 10. Revised by 김동욱
5. (New : Diagonalization) Let be a nondefective matrix with diagonalizing matrix . Show that the matrix diagonalizes .
Sol)
[ Show ]
#
#
, hence diagonalizes . ■
[ Note :
( ) 이므로,
]
2015.11.20 Solved by 김민기 2015.11.20. Revised by 남기현 2015.11.21 Revised by 박원기 2015.11.22 Finalized by 김도희 2015.11.23. Refinalized by 김민기 Final OK by SGLee
When we define the addition() and the scalar multiple() on and as follows. Check if and are vector spaces with the given two operations.
(1) .
(2) .
(3) , .
(4) , .
Sol)
(1) Check 10 laws of vector space conditions,
in general
is not a vector space.
(2) Check 10 laws of vector space conditions,
in general
is not a vector space.
(3) It satisfies 10 laws of vector space conditions.
is a vector space. ■
2015.11.30 Solved by 김민기 2015.11.30. Revised by 박원기 2015.11.30. Finalized by 김도희 Final OK by SGLee
Which one is a subspace of ?
(1) .
(2) .
(3) .
Solution)
if)
(1) : + not closed
So, by 2 step subspace test, (1) is not a subspace of .
(2) : + closed
: * closed
So, by 2 step subspace test, (3) is a subspace of .
(3) in general : not * closed
So, by 2 step subspace test, (4) is not a subspace of . ■
2015.11.20 Solved by 김민기 2015.11.20. Revised by 남기현 2015.11.21. Revised by 박원기 2015.11.22. Finalized by 김도희 2015.11.23. Refinalized by 김민기 Final OK by SGLee
Which one is a subspace of ?
(1) .
(2) .
(3) .
(4) .
Solution)
if)
(1) : + closed
: * closed
So, by 2 step subspace test, (1) is a subspace of .
(2) : + closed
: * closed
So, by 2 step subspace test, (2) is a subspace of .
(3) : + closed
: * closed
So, by 2 step subspace test, (3) is a subspace of .
(4) in general : not * closed
So, by 2 step subspace test, (4) is not a subspace of . ■
2015.11.20 Solved by 김민기 2015.11.20. Revised by 주민규 2015.11.21. Revised by 박원기 2015.11.22. Finalized by 김도희 2015.11.23. Refinalized by 김민기 Final OK by SGLee
Let be a vector in . Write as a linear combination of , , .
Sol) =
= , ()
=
■
2015.11.19. Solved by 2014312052 김도희 2015.11.20. Revised by 2014312749 남기현 2015.11.21. Revised by 2014313107 김민기 2015.11.23. Finalized by 2010310179 박원기 Final OK by SGLee
Determine if the given vectors in a given vector space are linearly independent or linearly dependent.
(1) .
(2) .
(3)
.
Sol)
(1) Let
and
(By using Sage)
http://math1.skku.ac.kr/home/pub/2528/
The vectors are linearly independent.
(2) Let
When , .
The vectors are linearly dependent.
(3)
When ,
The vectors are linearly dependent. ■
2015.11.30. Solved by 2014312052 김도희 2015.11.30. Finalized by 남기현 2015.11.30. Refinalized by 김민기 Final OK by SGLee
Let be the complex inner product space with the Euclidean inner product. Let . Answer the following.
(1) Compute .
(2) Compute .
(3) Confirm the Cauchy-Schwarz inequality.
(4) Confirm the triangle inequality.
Sol)
(1)
(2)
,
(3) , implies
1=
∴ Cauchy-Schwarz inequality holds.
(4) =>
∴ Triangle inequality holds. ■
2015.11.19. Solved by 2014312052 김도희 2015.11.20. Revised by 2014312749 남기현 2015.11.20.
Revised by 2014313107 김민기 2015.11.21. Finalized by 2010310179 박원기 Final OK by SGLee
Let be the complex inner product space with the Euclidean inner product. Let . Answer the following.
(1) Compute .
(2) Compute .
(3) Confirm the Cauchy-Schwarz inequality.
(4) Confirm the triangle inequality.
Sol)
(1) .
(2)
,
.
(3) , implies
1= =3.
(4) => .
∴ Triangle inequality holds. ■
2015.11.30. Solved by 김도희 2015. 11. 30. Revised by 배성준 2015.11.30. Revised by 박원기 2015.11.30. Revised by 김민기 Final OK by SGLee
Define an inner product on as = . Compute the following. (here = , = )
(1) The symmetric matrix A such that
(2) The norm of = (5, 0)
(3) The norm of = (-2, 4)
(4) such that
Sol)
(1)
Let ,
.
= .
by method of undetermined coefficients,
,
■
(2)
(3)
■
(4)
∴ ■
2015.11.18. solved by 박원기 2015.11.21. revised by 김도희 2015.11.22. Finallized by 김민기 2015.11.23. Final OK by TA LEE
Define an inner product on as = . Compute the following. (here = , = )
(1) The symmetric matrix A such that
(2) The norm of = (1, -1)
(3) The norm of = (4, 3)
(4) such that
Sol)
(1)
Let ,
.
.
by method of undetermined coefficients,
,
■
(2)
(3)
(4)
∴ ■
2015.11.18. solved by 박원기 2015.11.21. Revised by 이반석 2015.11.21. Revised by 김도희 2015.11.22. Finallized by 김민기 2015.11.23. Final OK by SGLee
Tell which one is a linear transformation or not. If not, give a reason.
(1)
(2)
(3)
(4)
(5)
(6)
Sol)
(1) 1.
2.
is a linear transformation.
(2) 1.
2.
is a linear transformation.
(3) 1.
2.
is a linear transformation.
(4) 1.
2.
is a linear transformation.
(5) 1.
is not a linear transformation.
(6) 1.
is not a linear transformation.
The cases (1),(2),(3),(4) indicate linear transformation. ■
2015.11.18. solved by 박원기 2015.11.21. revised by 김도희 2015.11.22. Finallized by 김민기 2015.11.23. Refinlized by 박원기 2015.11.24. Final OK by SGLee
Find the kernel and the range of the following linear transformation.
(1)
(2)
Sol)
(1)
{0}
Range of : = {}
(2) = = {}
For each , we can consider a constant function , so that
{}
■
2015.11.20 solved by 김민기
2015.11.20. revised by 정다산
2015.11.21. Revised by 박원기
2015.11.22. Finalized by 김도희
2015.11.23. Refinallized by 김민기
Final OK by SGLee
If are subspaces of a vector space , prove that is a subspace of .
Sol)
are subspaces of a vector space
Show is a subspace of .
Let
use the 2 step subspace test
∴
is a subspace of . ■
2015.11.20 solved by 김민기
2015.11.20. Revised by 주민규
2015.11.21. Revised by 박원기
2015.11.22. Finalized by 김도희
2015.11.23. Refinalized by 김민기
Final OK by SGLee
Let be a fixed vector and be the set of all vectors orthogonal to , that is, . Show that is a subspace of .
Sol)
Let ,
, .
, ( is a constant)
By the 2 step subspace test, is a subspace of .
■
2015.11.19. Solyed by 2014312052 김도희
2015.11.19. Finalized by 2014312749 남기현
2015.11.21. refinalized by 2014313107 김민기
2015.11.23. Refinalized by 2010310179 박원기
Final OK by SGLee
Let be the vector space with the Euclidean inner product. Transform into an
orthonormal basis by using the Gram-Schmidt process.
Sol)
[Step 1] Let
[Step 2] Let be a subspace spanned by and let
[Step 3] Let be a subspace spanned by and and let
by normalizing
∴orthonormal basis ■
-Solving problem by using Sage
http://math1.skku.ac.kr/home/pub/2529
x1=vector([I,I,I])
x2=vector([0,I,I])
x3=vector([0,0,I])
A=matrix([x1,x2,x3])
[G,mu]=A.gram_schmidt()
print G
B=matrix([G.row(i) / G.row(i).norm() for i in range(0, 3)]); B
Result
[ I I I]
[-2/3*I 1/3*I 1/3*I]
[ 0 -1/2*I 1/2*I]
[ 1/3*I*sqrt(3) 1/3*I*sqrt(3) 1/3*I*sqrt(3)]
[ -I*sqrt(2/3) 1/2*I*sqrt(2/3) 1/2*I*sqrt(2/3)]
[ 0 -I*sqrt(1/2) I*sqrt(1/2)]
2015.11.22. Solved by 김도희
2015.11.22. Revised by 정수호
2015.11.23. Revised by 김민기
2015.11.23. Finalized by 박원기
2015.11.30. Refinalized by 김도희
Final OK by SGLee
Find the standard matrix corresponding to the given linear transformation defined by
.
Sol)
When , where is the standard matrix and , we can write
.
Therefore, we now have equations ;
The standard matrix of is =.
■
2015.11.18. solved by 박원기
2015.11.18. revised by 김도희
2015.11.18. finalized by 정진규
2015.11.22. Refinalized by 김민기
2015.11.23. Refinalized by 김도희
2015.11.24. Final OK by SGLee
Define by Show that it is a linear transformation and find the kernel and the range of .
Sol)
for all .
for all .
Thus, by and , is a linear transformation.
,
Thus, = {}
{}={}
={}={}
={}
∴ set of all symmetric matrix in
■
2015.11.28. solved by 박원기
2015.11.30 revised by 김민기
2015.11.30 Finallized by 김민기
Show the set {} is a linearly independent vectors in a polynomial space composed of all of polynomial that have real coefficient and its degree is less than .
Sol)
If
for all ,
then (if not )
{} is a set of linearly independent vectors.
■
2015.11.28. solved by 박원기
2015.11.29. revised by 이반석
2015.11.30. revised by 김민기
If are linearly independent vectors, Show .
Sol)
,
.
,’s are linearly independent. so ,
.
■
2015.11.30 Solved by 김민기
2015.11.30. Revised by 주민규
2015.11.30. Revised by 박원기
2015.11.30. Finalized by 김도희
When we define the addition() and the scalar multiple() on and as follows. Check if and are vector spaces with the given two operations.
(1) .
(2) .
(3) , .
(1) Check 10 laws of vector space conditions,
in general
is not a vector space.
(2) Check 10 laws of vector space conditions,
in general
is not a vector space.
(3) It satisfies 10 laws of vector space conditions.
is a vector space. ■
2015.11.30 Solved by 김민기
2015.11.30. Revised by 박원기
2015.11.30. Finalized by 김도희
Which one is a subspace of ?
(1) .
(2) .
(3) .
Solution)
if)
(1) : + not closed
So, by 2 step subspace test, (1) is not a subspace of .
(2) : + closed
: * closed
So, by 2 step subspace test, (3) is a subspace of .
(3) in general : not * closed
So, by 2 step subspace test, (4) is not a subspace of .
■
2015.12.01. Solved by 2009313441 김동욱
2015.12.02. Revised by 이반석
2015.12.02. Finalized by 주민규, Final OK by SGLee
Determine if the given vectors in a given vector space are linearly independent or linearly dependent. (new)
(1) :
(2)
Sol)
(1) Let
and
(by using Sage) http://math1.skku.ac.kr/home/pub/2536 by 김동욱
A=matrix(QQ, 4, 4, [1,2,3,-2,2,-5,-1,3,3,-4,2,1,1,1,1,-1]) print A.det() |
The matrices are linearly independent. : OK
(2)
(by using Sage) http://math1.skku.ac.kr/home/pub/2537 by 김동욱
var('t') W=wronskian(t^4+t^3+t^2-t+3,t^4+2*t^3-3*t^2+t+4,2*t^4-2*t^3+5*t^2+8*t+1,-t^4+4*t^2+3*t-2) print W |
For some , . Thus these functions are linearly independent. (by Theorem 9.1.3) : OK■
2015.11.30. Solved by 2014312052 김도희 2015.11.30. Finalized by 남기현 2015.11.30. Refinalized by 김민기
Let be the complex inner product space with the Euclidean inner product.
Let . Answer the following.
(1) Compute .
(2) Compute .
(3) Confirm the Cauchy-Schwarz inequality.
(4) Confirm the triangle inequality.
Sol)
(1)
(2)
,
(3) , implies
1=
∴ Cauchy-Schwarz inequality holds.
(4) =>
∴ Triangle inequality holds.
■
2015.11.30. Solved by 김도희
2015. 11. 30. Revised by 배성준
2015.11.30. Revised by 박원기
2015.11.30. Revised by 김민기
Define an inner product on as = . Compute the following. (here = , = )
(1) The symmetric matrix A such that .
(2) The norm of = (5, 0)
(3) The norm of = (-2, 4)
(4) such that
Sol)
(1)
Let ,
.
= .
by method of undetermined coefficients,
,
■
(2)
(3)
■
(4)
∴ ■
Solved by 신영준
Finalized and Final OK by SGLee
Problem 1
Let be a 5×5 matrix with the only one eigenvalue with algebraic multiplicity of 5.
Find all possible types of Jordan Canonical forms of when the number of linearly independent eigenvectors corresponding is 2.
Sol)
Type 1
the dot diagram for is the following.
=2 : ⦁⦁
=1 : ⦁
=1 : ⦁
=1 : ⦁
So,
Type 2
the dot diagram for is the following.
=2 : ⦁⦁
=2 : ⦁⦁
=2 : ⦁
So, ■
Note: 5= 1+4, 2+3 (there are only 2 cases in the given decreasing order.)
Solved by 신영준 Finalized and Final OK by SGLee
Problem 2
For the given Jordan Canonical form , calculate the following:
(1)
Sol)
(2)
Sol)
(3)
Sol)
(4)
Sol) ■
Solved by 신영준, Finalized and Final OK by SGLee
Chapter 10, Problem 3
Find the Jordan Canonical Form of the given matrix.
Sol)
The matrix has its characteristic polynomial = .
So has two distinct eigenvalues , .
Here =25 has algebraic multiplicity 1, =0 has algebraic multiplicity 4.
Thus the dot diagram for has 1 dot.
=1 : ⦁ (number of Jordan block: 1)
Here has one Jordan block of 11. That is, =[25].
As well, the dot diagram for has 4 dot, and
= 5-rank(-) = 5-1 = 4
= rank() - rank[] = 1-1 =0
= rank[] - rank[] = 1-1 =0
= rank[] - rank[] = 1-1 =0
Thus, the dot diagram for is the following.
=4 : ⦁⦁⦁⦁ (number of Jordan block: 4)
has 4 Jordan blocks of 11 . That is, =.
Hence, the Jordan Canonical form of is
∴ ==.
Doble checked by Sage by 신영준. http://math1.skku.ac.kr/home/pub/2533 ■
Solved by 김선호, Revised by 신영준 Finalized and Final OK by SGLee
Chapter 10, Problem 4
Find the Jordan Canonical Form of the given matrix.
Sol)
The matrix has characteristic polynomial = .
So A has three distinct eigenvalues , , . Here =3 has algebraic multiplicity 1, =2 has algebraic multiplicity 1 and =1 has algebraic multiplicity 3.
Thus the dot diagram for has 1 dot.
=1 ; ⦁(number of Jordan block: 1)
Here has one 11 Jordan block. That is, =[3].
And also the dot diagram for has 1 dot.
=1 : ⦁(number of Jordan block: 1)
Here has one 11 Jordan block. That is, =[2].
As well, the dot diagram for has 3 dot, and
= 5-rank(-) = 5-4 = 1
= rank() - rank[] = 4-3 =1
= rank[] - rank[] = 3-2 =1
Thus, the dot diagram for is the following.
=1 : ⦁ (number of Jordan block: 1)
=1 : ⦁
=1 : ⦁
has one 33 Jordan block. That is, =.
Hence, the Jordan Canonical form of is
∴ ==
Doble checked by Sage by 신영준. http://math1.skku.ac.kr/home/pub/2539 ■
Solved by 김선호 Revised by 신영준 Finalized and Final OK by SGLee
Chapter 10, problem 5
Find the Jordan Canonical Form of the given matrix.
Sol)
The matrix has characteristic polynomial = .
So A has three distinct eigenvalues , , .
Here =5 has algebraic multiplicity 1, =1 has algebraic multiplicity 1 and =-3 has algebraic multiplicity 1.
Thus the dot diagram for has 1 dot.
=1 ; ⦁ (number of Jordan block: 1)
Here has one 11 Jordan block. That is, =[5].
And also the dot diagram for has 1 dot.
=1 : ⦁ (number of Jordan block: 1)
Here has one 11 Jordan block. That is, =[1].
As well, the dot diagram for has 1 dot, and.
=1 : ⦁ (number of Jordan block: 1)
has one 11 Jordan block. That is, =.
Hence, the Jordan Canonical form of is
∴ ==
Doble checked by Sage by 신영준. http://math1.skku.ac.kr/home/pub/2540 ■
Solved by 김선호, Revised by 이종현 finalized by 신영준, Final OK by SGLee
Problem 6 Find the Jordan Canonical Form of the given matrix.
Sol) The matrix has characteristic polynomial = .
So A has two distinct eigenvalues of , =2 and =0, each with algebraic multiplicity 2.
For =2,
= 4 - rank() = 4-2 = 2
= rank() - rank[] = 2-2 = 0
Therefore, the dot diagram for is the following.
=1 : ⦁ ( Number of Jordan blocks : 1)
=1 : ⦁
So, =
For =0
= 4 - rank = 4 - 3 = 1
= rank - rank[] = 3 - 2 = 1
=1 : ⦁ ( Number of Jordan blocks : 1)
=1 : ⦁
So, =.
Hence, the Jordan Canonical form of is
∴ == .
Doble checked by Sage by 김선호. http://math1.skku.ac.kr/home/pub/2534 ■
Solved by 이종현, Finalized and Final OK by SGLee
Problem7.
Find the Jordan Canonical Form of the given matrix.
Sol.
The matrix has characteristic polynomial = . So A has three distinct eigenvalues , , .
Here =4 has algebraic multiplicity 1, =2 has algebraic multiplicity 1 and =1 has algebraic multiplicity 3.
Thus the dot diagram for has 1 dot.
⦁
Here has 11 Jordan block. That is, =[4].
And also the dot diagram for has 1 dot.
⦁
Here has 11 Jordan block. That is, =[2].
Now the dot diagram for has 3 dot, and
= 5-rank(-) = 5-4 = 1
= rank() - rank[] = 4-3 =1
= rank[] - rank[] = 3-2 =1
Thus, the dot diagram for is the following. (number of Jordan block: 1)
=1 : ⦁
=1 : ⦁
=1 : ⦁
has ons 33 Jordan block. That is, =. Hence, the Jordan Canonical form of is
∴ == ■
Doble checked by Sage by 이종현. http://math1.skku.ac.kr/home/pub/2531
Solved by 이종현, Finalized and Final OK by SGLee
Problem8.
Find the Jordan Canonical Form of the given matrix.
Sol.
The matrix has characteristic polynomial = . So A has two distinct eigenvalues of , =4 and =2, each with alegebraic multiplicity 2.
For = 4,
= 4 - rank() = 4-3 = 1
= rank() - rank[] = 3-2 = 1
Therefore, the dot diagram for is the following. ( Number of Jordan blocks : 1)
=1 : ⦁ (Number of Jordan blocks = 1 )
=2 : ⦁
So, =.
For =2,
= 4 - rank() = 4-2 = 2
= rank() - rank[] = 2-2 = 0
Therefore, the dot diagram for is the following.
=2 : ⦁⦁ (Number of Jordan blocks = 2 )
So, =.
Thus, the Jordan Canonical Form of is =.
Doble checked by Sage by 이종현. http://math1.skku.ac.kr/home/pub/2532 ■ ]
Ch 10 Jordan Canonical Forms (Reference)
http://math1.skku.ac.kr/home/pub/2323/
http://matrix.skku.ac.kr/JCF/JordanCanonicalForm-SKKU.html
http://math1.skku.ac.kr/home/pub/656/
https://www.youtube.com/watch?v=NBLZPcWRHYI
MT Chapter 10 Jordan Canonical Forms
[ 1. Jordan Canonical Form
- Definition - Link
[ 2. Generalized Eigenvectors
- Definition - Link
- Finding a Jordan Canonical Form using Generalized Eigenvectors - Link
- Finding Generalized Eigenvectors using a Jordan Canonical Form - Link
[ 3. Characteristic Polynomial and Minimal Polynomial
- Characteristic Polynomial and Minimal Polynomial of a Jordan form - Link
- Diagonalization and Minimal Polynomial - Link
- Cayley-Hamilton theorem- Link
[ 4. Applications ]
- Computing J^k where J is a Jordan block - Link
- Computing A^k using Jordan Canonical Form - Link
- Computation of e^A using Jordan Canonical Form- Link
- Solving a Linear Ordinal Differential Equation - Link
1) The algebraic sum of currents in a network of conductors meeting at a point is zero.
3) Linear Algebra & Genetics. John Gallego, Sunit kambli, Daniel Lee http://web.csulb.edu/~jchang9/m247/m247_sp10_Danial_John_Sunny.pdf
http://www.math.washington.edu/~king/coursedir/m308a01/Projects/m308a01-pdf/kirkham.pdf
4) http://www.prenhall.com/esm/app/ph-linear/leon/html/proj1.html
https://en.wikipedia.org/wiki/Leslie_matrix
5) Casey Pearson. Matrices in Statics and Mechanics.(2012). http://home2.fvcc.edu/~dhicketh/LinearAlgebra/studentprojects/spring2012/Linear%20Algebra%20Project/Linear%20Algebra%20Project.pdf (2015.11.15.)
http://msemac.redwoods.edu/~darnold/math45/laproj/fall2006/iancraig/SVD_paper.pdf
7) http://www.sciencedirect.com/science/article/pii/0169743987800849
https://en.wikipedia.org/wiki/Principal_component_analysis
8) 대한전자공학회, http://www.ieek.or.kr/
“Low Complexity Power Allocation Scheme for MIMO Multiple Relay System With Weighted Diagonalization”
9) W.C.Chew, S.Koc, J.M. Song, C.C.Lu, and E. Michielssen / Center for computational electromagnetics / Electromagnetics Laboratory / Department of electrical and Computer Engineering / University of Illinois / Urbana, Il 61801
http://scholar.ndsl.kr/schDetail.do?cn=NPAP00247997
Made by SGLee http://matrix.skku.ac.kr/sglee/