Chapter 8. Eigenvalues and Eigenvectors
2018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: E12 (금 12:00 - 14:45), 반도체관 400126호 담당교수: 김응기 박사
주차 |
주교재 |
부교재 |
6 |
8.1: Eigenvalues and Eigenvectors 8.4: Eigenbases, Diagonalization , Quadratic Forms |
1.6 고유값과 고유벡터 1.7 닮음, 행렬의 대각화, 이차형식 |
web |
Week 6
8.1 Eigenvalues, Eigenvectors
From the view point of engineering application, eigenvalue problems are among the most important problems in connection with matrices.
Let be a given matrix and consider the vector equation
(1) .
Here is an unknown vector and is an unknown scalar.
Our task is to determine ’s and ’s that satisfy .
should be proportional to .
The zero vectors is a solution of for any value of , because .
A value of for which has a solution is an eigenvalue ( characteristic value) of the matrix .
The corresponding solutions of are an eigenvectors (characteristic vectors) of corresponding to that eigenvalue .
The set of all the eigenvalues of is called the spectrum of .
The spectrum consist of at least one eigenvalue and at most numerical different values.
The largest of the absolute values of eigenvalues of is called the spectral radius of .
How to Fine Eigenvalues and Eigenvectors
is homogeneous linear system
has a non-trivial solution is zero.
: Characteristic determinant or characteristic polynomial
: characteristic equation of
The eigenvalues of are the solutions of the characteristic equation of .
Example 1 Determination of Eigenvalues and Eigenvectors
The eigenvalues and eigenvectors of the matrix .
Solution
Eigenvalues
: characteristic determinant, characteristic polynomial
: characteristic equation of
Solutions of this quadratic equation are , .
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
.
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector
.
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
① Characteristic polynomial of
Evaluate
x^2 + 7*x + 6
② Hence the eigenvalues are as follows.
Evaluate
[x == -6, x == -1]
③ We can find the eigenvalues directly by using the built in command.
Evaluate
[-1, -6]
④ In order to find eigenvector for , solve
Evaluate
[ 2 -1]
[ 0 0]
We obtain the eigenvector
.
⑤ In order to find eigenvector for , solve .
Evaluate
[1 2]
[0 0]
We obtain the eigenvector
.
⑥ We can find the eigenvectors directly by using the built in command.
Evaluate
[(-1, [(1, 2)], 1), (-6, [(1, -1/2)], 1)]
Equation written in components is
Transferring the terms on the right side to the left, we have
In matrix notation
(3)
This homogeneous linear system of equations has a nontrivial solution if and only if the corresponding determinant of the coefficients is zero :
(4)
is the characteristic matrix.
is the characteristic determinant of .
is the characteristic equation of .
By developing we obtain a polynomial of th degree in .
This is the characteristic polynomial of .
Theorem 1 Eigenvalues
The eigenvalues of a square matrix are the roots of the characteristic equation (4) of . Hence an matrix has at least one eigenvalue and at most numerically different eigenvalues.
Theorem 2 Eigenvectors, Eigenspace
If and are eigenvectors of a matrix corresponding to the same eigenvalue ,
so are and for any .
Hence the eigenvectors corresponding to one and the same eigenvalue of , together with , from a vector space called the eigenspace of corresponding to that .
Proof
and imply
and
hence
In particular, an eigenvector is determined only up to a constant factor.
Hence, we can normalize , that is, multiply it by a scalar to get a unit vector.
For instance
Let
Length of :
Normalized of :
Sage Coding
Example 2 Multiple Eigenvalues
Find the eigenvalues and eigenvectors of
.
Solution
The characteristic determinant gives the characteristic equation
.
The roots (eigenvalues of ) are , (Double root).
For the characteristic matrix is
Hence it rank .
.
Choosing we obtain form .
Taking and , we obtain from .
Hence an eigenvector of corresponding to is .
For the characteristic matrix is
.
Hence it rank .
.
Choosing , .
Choosing , .
We obtain two linearly independent eigenvectors of corresponding to ,
and .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[(5, [(1, 2, -1)], 1), (-3, [(1, 0, 1/3),(0, 1, 2/3)], 2)]
Algebraic multiplicity of
The order of an eigenvalue as a root of the characteristic polynomial is called the algebraic multiplicity of .
Geometric multiplicity of
The number of linearly independent eigenvectors corresponding to is called the geometric multiplicity of .
Thus is the dimension of the eigenspace corresponding to this . Since the characteristic polynomial has degree , the sum of all the algebraic multiplicities must equal .
In general .
Defect of
The difference is called the defect of .
In example 2
For : and .
Example 3 Algebraic multiplicity, Geometric multiplicity, Positive defect
The characteristic equation of the matrix
is .
(double root)
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
If we choose ,
, .
We obtain the eigenvector
.
Algebraic multiplicity is .
Geometric multiplicity is ,
Defect is .
Since eigenvectors result from , hence , in the from .
Hence for the defect is .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[(0, [(1, 0)], 2)]
The characteristic equation of the matrix
is .
(double root)
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
An eigenvalue of algebraic multiplicity . but its geometric multiplicity is only , since eigenvectors result from in the from .
For , the defect is .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[(3, [(1, 0)], 2)]
Example 4 Real Matrices with Complex Eigenvalues and Eigenvectors
Since real polynomials may have complex roots, and real matrix may have complex eigenvalues and eigenvectors.
The characteristic equation of the skew-symmetric matrix
is .
Two eigenvalues , .
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[(-1*I, [(1, -1*I)], 1), (1*I, [(1, 1*I)], 1)]
Theorem 3 Eigenvalues of the Transpose
The transpose of a square matrix has the same eigenvalues as .
Proof
Transposition does not change the value of the characteristic determinant.
8.4 Eigenbases. Diagonalization. Quadratic Forms
General properties of eigenvectors
Eigenvectors of an matrix may form a basis for . If we are interested in a transformation , such an “eigenbasis”(basis of eigenvectors).
We can represent any in uniquely as a linear combination of the eigenvectors , , , ,
.
And denoting the corresponding eigenvalues of the matrix by , , , , we obtain
(1)
Theorem 1 Basis of Eigenvectors
If an matrix has distinct eigenvalues, then has a basis of eigenvectors , , , for .
Example 1 Eigenbases. Nondistinct Eigenvalues. Nonexistence
The eigenvalues and eigenvectors of the matrix .
Solution
Eigenvalues
Characteristic equation is
Roots (eigenvalues of ) are , .
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
Eigenvector of corresponding .
Solution is .
This determines an eigenvector corresponding to .
We choose ,
, .
Eigenvector is
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[(8, [(1, 1)], 1), (2, [(1, -1)], 1)]
Even if not all eigenvalues are difference, a matrix may still provide an eigenbasis for .
A may not have enough linearly independent eigenvectors to make up a basis. For instance
and has only one eigenvector ( arbitrary).
Theorem 2 Symmetric Matrices
A symmetric matrix has an orthonormal basis of eigenvectors for .
Example 2 Orthonormal Basis of Eigenvectors
The matrix is symmetric, and an orthonormal basis of eigenvectors is
, .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[ 1 1]
[ 1 -1]
[ 0.7071067811865475? 0.7071067811865475?]
[ 0.7071067811865475? -0.7071067811865475?]
Similarity of Matrices, Diagonalization
Definition Similar Matrices Similarity Transformation
An matrix is similar to an matrix if
for some matrix .
This transformation, which gives from is a similarity transformation.
.
Theorem 3 Eigenvalues and Eigenvectors of Similar Matrices
If is similar to , then has the same eigenvalues as . Furthermore, if is an eigenvector of , then is an eigenvector of corresponding to the same eigenvalue.
Proof
( an eigenvalue, ) we get .
Now .
By this “identity trick” the previous equation gives
.
Hence is an eigenvalue of and a corresponding eigenvector.
Indeed, would give , contradicting .
Example 3 Eigenvalues and Eigenvectors of Similar Matrices
Let and .
Then .
Here was obtained with . We see that has the eigenvalues , .
Characteristic equation of is .
The roots (the characteristic equation of ) is , .
From the first component of we have .
For this gives , say, .
For this gives , say, .
We have
,
These are eigenvectors of the diagonal matrix . We see that and are the column of . This suggests the general method of transforming a matrix to diagonal from by using , the matrix with eigenvectors as column.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
([3 0] [ 1 1] ,[0 2], [ 1 4/3] )
Theorem 4 Diagonalization of Matrix
If an matrix has a basis of eigenvectors, then
(5)
is diagonal, with the eigenvalues of as the entries on the main diagonal. Here is the matrix these eigenvectors as column vectors. Also
()
Example 4 Diagonalization
Diagonalize .
Solution
The characteristic equation is .
The roots (eigenvalues of ) are , , .
By the Gauss elimination applied to with .
We find eigenvectors and then by the Gauss-Jordan elimination.
The results are
, , , ,
Calculating and multiplying by from the left, we thus obtain
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
True
[(3, [(1, -3, 1)], 1), (0, [(1, 1/2, 2)], 1), (-4, [(1, -1, 3)], 1)]
Evaluate
[ 3 0 0]
[ 0 0 0]
[ 0 0 -4]
Quadratic Forms. Transformation to Principal Axes
By definition a quadratic forms in the components , , , of a vector is a sum of terns, namely,
(7)
is called the coefficient matrix of the from. The matrix is symmetric.
We can take off-diagonal terns together in pairs in pairs and write the result as a sum of two equal terns.
Example 5 Quadratic Form. Symmetric Coefficient Matrix
Let .
Here .
From the corresponding symmetric matrix , where ,
thus , , , we get the same result,
.
By Theorem 2 the symmetric coefficient matrix of has an orthonormal basis of eigenvectors. Hence if we take these as column vectors, we obtain a matrix that is orthogonal.
().
. (8)
If we get , then, since , we get
. (9)
Furthermore, in (8) we have and , so that becomes simply
. (10)
Theorem 5 Principal Axes Theorem
The substitution transforms a quadratic form
()
to the principal axes form or canonical form , where , , , are the (not necessarily distinct) eigenvalues of the matrix , and is an orthogonal matrix with
corresponding eigenvectors , , , , respectively, as column vectors.
Example 6 Transformation to Principal Axes. Conic Sections
Find out what type of conic section the following quadratic form represents and transform it to principal axes :
.
Solution
We have , we have
, .
The characteristic equation .
The roots (eigenvalues of ) are , .
Hence (10) becomes
.
represents the ellipse , that is
.
The direction of the principal axes in the -coordinates,
we have determine normalized eigenvectors from with and
and then (9). We get
and
hence
, .
This is rotation.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
① Computing eigenvalues of
Evaluate
[32, 2]
② Computing eigenvectors of
Evaluate
[(32, [(1, -1)], 1), (2, [(1, 1)], 1)]
③ Computing diagonalizing
Evaluate
[ 1/2*sqrt(2) 1/2*sqrt(2)]
[-1/2*sqrt(2) 1/2*sqrt(2)]
④ Sketching two ellipses simultaneously
Evaluate
http://matrix.skku.ac.kr/sglee/
[한빛 아카데미] Engineering Mathematics with Sage:
[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화
Contents
A. 공학수학 1 – 선형대수, 상미분방정식+ Lab
Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html
B. 공학수학 2 - 벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html
Copyright @ 2018 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).