SKKU-LA-CH-3-SGLee


                        

 

 

Chapter 3

 

 

 

 

Matrix and Matrix Algebra

 

 3.1 Matrix operation

 3.2 Inverse matrix

 3.3 Elementary matrix

 3.4 Subsapce and linear independence

 3.5 Solution set of a linear system and matrix

 3.6 Special matrices

*3.7 LU-decomposition

 

Matrix is widely used as a tool to transmit digital sounds and images through internet as well as solving linear systems. We define the addition and product of two matrices.

These operations are tools to solve various linear systems. Matrix product also becomes an excellent tool in dealing with function composition.

 

In the previous chapter, we have found the solution set using the Gauss elimination method.

In this chapter, we define the addition and scalar multiplication of matrices and introduce algebraic properties of matrix operations.

It will be used to describe the relation between solution set and matrix, Then using the Gauss elimination, we show how to find the inverse matrix. 


Furthermore, we investigate the concepts such as linearly independence and subspace which are necessary in understanding the structure of a linear system.

Finally we introduce some interesting special matrices.

 

 

 

3.1 Matrix operation

 

 Reference video: https://youtu.be/C56kVi-AZW8 (http://youtu.be/DmtMvQR7cwA)  

 Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-3-Sec-3-1.html 

 

 

 

 

This chapter introduces the definition of the addition and scalar multiplication of matrices and the algebraic properties of the matrix operations.  Although many of the properties are identical to those of the operations on real numbers, some properties are different. Matrix operation is a generalization of the operation on real numbers.

 

Definition

 [Equality of Matrices]

 

 

 

 

Two matrices  and  of same size are equal if  for all , and denote it by .

 

 

 

 


 To define equal matrices, the size of two matrices should be the same.

 

 

For what values of  the two matrices

 

 

are equal?

 

 For each entry should be equal. Thus (that is, .  

 

 

Definition

 [Addition and scalar multiplication of matrix]

 

 

 

 

Given two matrices  and  and a real number , the sum  of  and , and the scalar multiple  of  by  are defined by

.

 

 

 

 

 

 To define addition, the size of two matrices should be the same.

 

 

For   , what is ?

 

 ,

       

    .                □

● http://matrix.skku.ac.kr/RPG_English/3-MA-operation.html 

● http://matrix.skku.ac.kr/RPG_English/3-MA-operation-1.html 

http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

 




                                                                  

[ 1  3  0]            [ 2  4 –8]              [-1 -1]

[-3  4  4]            [-4  2  6]              [-2 –2]            

 

 

Definition

 [Matrix product]

 

 

 

 

Given two matrices  and , the product  of  and  is defined below.

,

where  .

 

 

 

 

 

For two matrices  and  to be compatible for multiplication, we require the number of columns of  to be equal to the number of rows of .

The resultant matrix  is of size number of rows of  by the number of columns of .

 

[Remark]

 

 

 

 

 

 

 

 

 

[Remark]

Meaning of matrix product

 

 

 

 

 Let  , and denote the th row of  by  and the th column of  by . Then

Thus, 

Note that the inner product of th row vector of  and the th column vector of  is the  entry of .

 

[King Sejong's '' rule] 

 

 

 

 

 

 

Let . Then     


      □

 

● http://matrix.skku.ac.kr/RPG_English/3-MA-operation-1-multiply.html

 

 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

 




 Using matrix product, one can express a linear system easily. Let us consider the following linear system

 

 

          

          

                            

          

 

   and let

            

  be the coefficient matrix, the unknown vector and the constant vector respectively. 

Then we can express the linear system as          

                 

Theorem

 3.1.1

Let  be matrices of proper sizes (oeprations are well defined) and let  be scalars. Then the following hold.

 (1)               (commutative law of addition)

 (2)   (associative law of addition)

 (3)            (associative law of multiplication)

 (4)        (distributive law)

 (5)        (distributive law)

 (6) 

 (7) 

 (8) 

 (9) 

 

The proof of the above facts are easy and readers are encouraged to prove them.

 

Check the associative law of the matrix product.

 Since , we have

       

       Since , we have

        Hence, .

       

 

 

 

 The properties of operations on matrices are similar to those of operations on real numbers which are well known,

 Exception: For matrices , we do not have  in general.

 

 

 

Suppose that we are given the following matrices .

.

Then  is defined but  is not defined. Similarly  is a matrix but  is a  matrix, and hence . Also although  and  are  matrices, as we can see below, we have .

                                  

 

 

 

[Remark]

Computer simulation

 

 

 

 

  [matrix product] (Commutative law does not hold.)

  http://www.geogebratube.org/student/m12831

 

 

 

 

 

 

 

Definition

 [Zero matrix]

 

 

 

 

A zero matrix consists of entries of 0's and denoted by (or ).

 

 

 

 

 

Theorem

 3.1.2

For any matrix  and a zero matrix  of a proper size, the following hold.

 (1) 

 (2) 

 (3) 

 (4) 

 

 Note: Although , it is possible to have . Similarly,

        although , it is possible to have .

 

 

Let   .

Then . But  and .

Also  but .                                 

 

 



We should first define scalar matrices.

 

Definition

 [Identity matrix]

 

 

 

 

A scalar matrix of order  with diagonal entries all 1's is called an identity matrix of order  and is denoted by . That is,

 

 

 

 

 Let  be an  matrix and the identity matrix . It is easy to see that

 

                                                .


 

Let . Then ,

.

 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

 





[ 4 -2  3]

[ 5  0  2]

 

[ 4 -2  3]

[ 5  0  2]

 

[0 0]

[0 0]                                                            ■

 

Definition

 

 

 

 

 

Let  be a square matrix of order . The th power of  is defined by

( times)

 

 

 

 

 

Theorem

 3.1.3

If  is a square matrix and  are non negative integers, then

.

 

Let . Find  and confirm that .

http://sage.skku.edu or http://mathlab.knou.ac.kr:8080

 




[  6  -8]

[ 20 -10]

 

[-16 -12]

[ 30 -40]

 

[1 0]

[0 1]

 

True                                                             ■

  In the set of real numbers, we have  .

      However, the commutative law under matrix product does not work and thus we only have the following.

 

                                       .

 

   When , we have .

 

 

Definition

 [Transpose matrix]

 

 

 

 

For a matrix , the transpose of  is denoted by and defined by

.

 

 

 

 

 

 The transpose  of  is obtained by interchanging the rows and columns of .

 

Find the transpose of the following matrices.

  ,  

 ,

       .                                □

 http://sage.skku.edu 또는

 http://mathlab.knou.ac.kr:8080




 1  4]                   [ 5 –3  2]               [3]

[-2  5]                [ 4  2  1]               [0]

[ 3  0]                                              [1]               ■

 

Theorem

 3.1.4

Let  be matrices of appropriate sizes and  a scalar. The following hold.

 (1) 

 (2) 

 (3) 

 (4) .

 

Let . Show that (3) of Theorem 3.1.4 is true.

 Since .

Also, .  Thus .

                                                                  

 

 

Definition

 [Trace]

 

 

 

 

The trace of  is defined by  .

 

 

 

 

 

Theorem

 3.1.5

If  are square matrices of the same size and , then

 (1) 

 (2) 

 (3) 

 (4) 

 (5) 

 We prove the item (5) only and leave the rest as an exercise.

              .     

 

Let . Show that (5) of Theorem 3.1.5 is true.

 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080




37

 

37                                                                ■

3.2 Inverse matrix

 Reference video: https://youtu.be/naFiYy4RTxA (http://youtu.be/GCKM2VlU7bw)    

 Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-3-Sec-3-2.html 


  In this chapter, we introduce an inverse matrix of a square matrix which plays like a multiplicative inverse of a real number.

 We investigate the properties of an inverse matrix.

  You will see that some properties holding in the inverse of a real number are not true in the matrix inverse operation although most hold in both inverses.


Definition

 

 

 

 

 

A square matrix  of order  is called invertible (or nonsingularif there is a square matrix  such that

.

This matrix  if exists is called the inverse matrix of . If such a matrix  does not exist,  is called noninvertible, (orsingular).

 

 

 

 


Let . Note that the third row of  has all zeroes. Thus for any matrix

 the third row of  is . Therefore there does not exist  such that , that is,  is singular.




False                                                           ■

 

Theorem

 3.2.1

If  is an invertible square matrix of order , then an inverse of  is unique.

 Suppose that  are inverses of . Then as

 

                     

we have

                             

Thus an inverse of  is unique.                                          


A necessary and sufficient condition for  to be invertible is that . Hence one has

 

                         .

It is straightforward to check

        


Theorem

 3.2.2

If  are invertible square matrices of order  and  is a nonzero scalar, then the following hold.

(1)  is invertible and .

(2)  is invertible and .

(3)  is invertible and .

(4)  is invertible and .

  (2)    .

       (3)~(4) Just check that the product of matrices are the identity matrix.      

 

 

Theorem

 3.2.3

If  is an invertible matrix, then so is  and the following holds.

.

               .                               

 

 

Let Check that .

 Since ,

       , we have

  . Also since

   we have

 .                

● http://matrix.skku.ac.kr/RPG_English/3-SO-MA-inverse.html 




3.3 Elementary matrices

 

 

 Reference video: https://youtu.be/pcnFDa8K8ZY (http://youtu.be/GCKM2VlU7bw) 

 Practice site:  http://matrix.skku.ac.kr/knou-knowls/CLA-Week-3-Sec-3-3.html 

 


In the previous section, we defined an inverse of square matrices.

In this section, we shall discuss how to find an inverse of square matrices by using elementary row operations and elementary matrices.

 

 

Definition

 

 

 

 

 

An  by   matrix is called an elementary matrix if it can be obtained from  by performing a single elementary row operation (ERO). permutation matrix is obtained by exchanging rows of .

 

 

 

 

 

 

Listed below are three type of elementary matrices (Type 1, 2, 3) and the operations that produce them.

 : Interchange the 2nd and the 3rd rows.    

 : Multiply the 2nd row by 3.                 

 : Add 2 times the 1st row to the 2nd row.  




[1 0 0 0]          [ 1  0  0  0]          [1 0 0 7]

[0 0 1 0]          [ 0  1  0  0]          [0 1 0 0]

[0 1 0 0]          [ 0  0 -3  0]          [0 0 1 0]

[0 0 0 1]          [ 0  0  0  1]          [0 0 0 1]               ■

[Property of elementary matrix] The product of an elementary matrix  on the left and any matrix  is the matrix that results when the corresponding same row operation is performed on .

 

                  [Type 1] 

                   [Type 2]  

                 [Type 3]  

[Property of elementary matrix] The product of an elementary matrix  on the left and any matrix  is the matrix that results when the corresponding same row operation is performed on .

 

   [Type 1] 

     [Type 2]  

 [Type 3]  




[1 2 3]          [1 2 3]          [1 2 3]

[0 1 3]          [3 5 7]          [3 3 3]

[1 1 1]          [0 1 3]          [0 1 3]                         ■

[Remark]   The inverse of an elementary matrix is elementary.

Since           [Type 1]

 Since   [Type 2]

  Since   [Type 3]





 [1 0 0]         [  1   0   0]       [ 1  0  0]        

 [0 0 1]         [  0 1/3   0]       [ 0  1  0]        

 [0 1 0]         [  0   0   1]       [ 0 -4  1]

Finding the inverse of an invertible matrix.

 

We investigate the method to find the inverse of an invertible matrix using elementary matrices.

First consider equivalent statements of an invertible matrix (its proof will be treated in Chapter 7).

 

Theorem

 3.3.1 [Equivalent statements]

For any  matrix , the followings are equivalent.

(1)  is invertible.

(2)  is row equivalent to . (i.e. RREF)

(3)  can be expressed as a product of elementary matrices.

(4)  has only the trivial solution .

 

[Remark]

 

 

 

 

 

              

 

 

 

 

 

 

Theorem

 3.3.2 [Computation of an inverse]

 

[Remark]

Finding an inverse using the Gauss-Jordan elimination.

 

 

 

 

  [Step 1] For a given , augment  on the right side so that we make

         a  matrix .

  [Step 2] Compute the RREF of .

  [Step 3] Let  be the RREF of  in the step 2. Then,

 following hold.

        (ⅰ) If , then .

        (ⅱ) If , then  is not invertible so that  does not exist.

 

 

 

 

 

 

Find the inverse of 

 Consider . Then

                              

and, its RREF is given as follows.

      

 

 

Since .

 

         ∴                              

 

 

 

 

 

Find the inverse of

 It follows from a similar way to Example 03,

    

 

Since  does not exist. .                     

 

 

Find the inverse of

                                

● http://matrix.skku.ac.kr/RPG_English/3-MA-Inverse_by_RREF.html

  




[     1      0      0  |  8/15 -19/15   2/15]

[     0      1      0  |  1/15 -23/15   4/15]

[     0      0      1  |  4/15 –2/15    1/15]

We can extract inverse of  using slicing of the above matrix.

 

Aug[:, 3:6]

                                                                   

[  8/15 -19/15   2/15]

[  1/15 -23/15   4/15]

[  4/15  -2/15   1/15]

 

Thus .                                         

3.4 Subspaces and Linear Independence

 

 Reference video: https://youtu.be/bFh4MM9sJek(http://youtu.be/HFq_-8B47xM)     

 Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-4-Sec-3-4.html

In this section, we define a linear combination, a spanning set, a linear (in)dependence and a subspace of .

We will also learn how to solve the system of linear equations by using the fact

that solutions for a system of homogeneous linear equations form a subspace of .

 

 

* Note that  with standard addition and scalar multiplication is also called a vector space over  and its elements are called vectors.

 

Definition

 [Subspace]

 

 

 

 

Let  be a nonempty subset of . Then  is called a subspace of  if  satisfies the following two conditions.

 

   (1)    (closed under the addition)

   (2)   (closed under the scalar multiplication)

 

 

 

 

 

 All subspaces of  contain zero vector.

                      


 

 and  are subspaces of  where  is denoted by the origin. They are called the trivial subspaces.

 

 

 

 

 

A subset  of  satisfies two conditions for subspace. Hence,  is a subspace of . On the other hand, a subset  of  does not satisfy conditions for subspace so that  is not a subspace of .

 

       ,  but          

 

 

 

 

All subspaces of  are one of the followings.

1. zero subspace : 

2. Lines through the origin.

3. 

All subspaces of  are one of the following.

1. zero subspace : 

2. Lines through the origin

3. Planes through the origin

4.                                                         

 

 

 

 

 

 

Show that a subset  is a subspace of.

 

 For

the following hold.

(ⅰ) 

(ⅱ) 

Therefore,  is a subspace of .                    

 

 

 

 

Let  denote the set of all  matrices over .

 

 

For , show that

is a subspace of . (This  is called a solution space or null spaceof )

 

 Clearly,  so that  and . Since for 

,

we can obtain that

 and

                                  .

      This implies  and .

Therefore,  is a subspace of .                   

 

 

 

 

Definition

 [linear combination]

 

 

 

 

If  can be expressed in the form

 

 

with , then  is called a linear combination of vectors .

 

 

 

 

 

 

Let  be vectors of . Can  be a linear combination of and ?

 

 The answer is depend on whether there exist   in  such that

 

                                     .

From this observation, we can obtain

                 

 

   One can easily show that the above system has no solution.




[1 0 0]

[0 1 0]

[0 0 1]

 

Since this system of linear equation has no solution, there are no such scalars  exist. Consequently,  is not a linear combination of  .                                                      

 

Show that the set of all linear combinations of 

is a subspace of .

 

 Let . Then there exist  such that

.

Hence

,

                 and .

      This implies .

Hence,  is a subspace of                          

 

 

 

   In , we saw that for a subset  ,

     the set of all linear combinations  of  is a subspace of .

We say  is a subspace of  spanned by . In this case, we say  spans  and S is a spanning set of . We denote it

 

                     or .

  In particular, if all vectors in  can be expressed a linear combination of , then  spans . That is,

 

                       

 

  

 

 (i) Show that  is a spanning set of .

 (ii) Show that  is a spanning set of .

 

 

 

 

 

Definition

 [column space and row space]

 

 

 

 

Let . Then,  columns  of  span a subspace of . This subspace is called a column space of , denote by

 or Col().

  Similarly, a row space of  is defined by a subspace of  spanned by  rows  of , denoted by

 or Row().

 

 

 

 

 

For

determine whether  spans  or not.

 

 This is a question whether there exist  such that a given vector  is written as

 

(Using column vectors)

                   




[1 0 1]

[0 1 1]

[0 0 0]

 

This means that one of  cannot be determined. Therefore this linear system has a case that the system cannot determine a unique solution.                                               

 

Definition

 [Linearly Independent and Linearly Dependent]

 

 

 

 

If  satisfies

                    

then (or subset ) are called linearly independent.

If (or subset ) are not linearly independent, then it is called

linearly dependent.

 

 

 

 

 

  If  is linearly dependent, there exist at least one non-zero scalar

in  such that

 

                      .

 

 

 

 The unit vectors of 

                                     

 

 are linearly independent. This is because

                   

                

                

                .

 

Show that for  is linearly independent.

 

 For any ,

                     

                                  

Thus , and  is linearly independent.               

 

Show that if  in  are linearly independent, then

are also linearly independent.

 

 For any ,

     .

Since  are linearly independent,

                             

Therefore  are linearly independent. 

 

 

 

 

For        

in , Show that  is linearly dependent.

 

  For any , if , then

              




[ 1  0 -1]

[ 0  1  1]

[ 0  0  0]

 

This means that the above equations can be reduced to two equations of three variables. Since it has three variables more than the number of equations so that there are non-trivial solutions. One of them is given by . Therefore there exist non zero scalars  is linearly dependent.   

Theorem

 3.4.1

For a set , the followings hold.

(1) A set  is linearly dependent if and only if some element in  can be expressed as a linear combination of the other elements in .

(2) If  contains the zero vector, then  is a linearly dependent.

(3) If a subset  is linearly dependent, then  is also linearly dependent.

   If  is linearly independent, then  is also linearly independent.

 

 (1) () If  is linearly dependent, then there exist  such that

 

                                             

where at least one element in  is a nonzero.

Without loss of generality, if  then,

 

                         

so that  can be expressed as a linear combination of the other vectors in 

() Without loss of generality, we can write

 

                             

so that

                          

Hence,  is linearly dependent since .

 

Proofs of the rest are left as an exercise.                

 

 In other words, that set  is linearly independent means that any vector in  cannot be written as a linear combination of the other vectors in .


 In , there are at most  vectors in a linearly independent set.

 

Theorem

 3.4.2 (For proof, see Theorem 7.1.2)

In  vectors are always linearly dependent.

 

 

For  in , we can easily check that  is linearly dependent from Theorem 3.4.2.                     

 

 

 

 

[Remark]

Lines and planes (from the viewpoint of subspace)

 

 

 

 

 (1) Note that the span of nonzero vector  in  is a subspace containing the zero vector. Also  forms a line through  and parallel to . In other words,  is translate of  by .

 (2) In general, if  are vectors in , then the set of vectors () is a subset of  which is the translation of a subspace that pass through the origin, by .

 

 

 

 

 

 

 

 

 

3.5 Solution set and matrices

 

 

 Reference video: https://youtu.be/E9HHrchqXus (http://youtu.be/daIxHJBHL_g )

 Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-4-Sec-3-5.html


 

 

 

 

In this section, we first state the relationship between invertibility of matrices and solutions to systems of linear equations, and then consider homogeneous systems.


 

Theorem

 3.5.1 [Relation between an invertible matrix and its solution]

If  an  matrix  is invertible and  is a vector in , the system

has a unique solution .

 

The following system can be written as .

 

                        

where . It is easy to show that  is invertible,

          and .

Thus the solution of the above system is given by

 

           .

That is .                               




x= (-1, 1, 0),     x= (-1, 1, 0)                                  ■

[Remark]

The homogeneous linear system

 

 

 

 

 

 

  

 

can be written as , where

 

     

 

  The vector  is called a trivial solution, and the solution  is called a nontrivial solution. Since a homogeneous linear system always has a trivial solution, there are two cases as follows.

 

  (1) It has only a trivial solution.

  (2) It has infinitely many solutions (i.e. it has nontrivial solutions as well.)

 

 

 

 

 

 

 

Theorem

 3.5.2 [Nontrivial solution of a homogeneous system]

A homogeneous system with equations and  variables such that 

 (i.e. the number of variables is greater than that of equations) has nontrivial solutions.

 

 

Since the existence of multiple solutions (provided that there is any solution at all) depends only on

the coefficient matrix and since a homogeneous system always has at least one solution (namely the trivial one),

multiple solutions for a linear system are possible only if the corresponding homogeneous system has multiple solutions.

But the homogeneous system has multiple solutions if and only if it has a non-trivial solution.

 

The homogeneous linear system

 

                  

has the following augmented matrix and its RREF.




A=

[1 1 1 1 0]

[1 0 0 1 0]

[1 2 1 0 0]

 

RREF(A)=

[ 1  0  0  1  0]

[ 0  1  0 -1  0]

[ 0  0  1  1  0]

 

The corresponding system of equations is

 

                             

Let (: a real number). Then the solution to (2) is

 

                      .

The solution is trivial if , and nontrivial if .       

 

Definition

 [The associated homogeneous system of linear equations]

 

 

 

 

Given a linear system  is called the associated homogeneous system of linear equations of .

 

 

 

 

 

Consider a system of linear equations.

 

                     

The associated homogeneous linear system is as the following:

 

                                     

 Since the matrix size is greater than 2, let us use Sage.

 The RREF of the augmented matrix of the above system is as follows :




[  1   0   0   4   2   0   0]

[  0   1   0   0   0   0   0]

[  0   0   1   2   0   0   0]

[  0   0   0   0   0   1 1/3]

 

Thus the above system reduces to

 

            .

 

Note that  and  are free variables.

Let . Then we have

 

                       .

 

Consider the augmented matrix of RREF of its associated homogeneous linear system.

 




[1 0 0 4 2 0 0]

[0 1 0 0 0 0 0]

[0 0 1 2 0 0 0]

[0 0 0 0 0 1 0]

 

It is easy to see that the solution to this system is given by

 

                                          .           

 

When compared geometrically the solutions to a system and

those of an associated homogeneous linear system,

the solution set for the associated homogeneous linear system is

merely translated by the vector  below.

 

                          

  We call the vector  a particular solution which can be obtained by substituting .

[Remark]

 

Relation between the solution set of the linear system and that of the associated homogeneous linear system.

 

 

 

 

  If  and , then

.

 Thus a system of linear equation  has solutions. Let  be a solution space to . If  is a solution to , then

is a solution set of .

 

 

 

 

 

 

 

 

 

 

 and 

 

 

 

 

 

 

 

 

 A geometric meaning of  which is a solution set of  is a set of translation 

when a particular solution  is added to a solution set  of .

 

      Since  does not contain a zero vector, it is not a subspace of .


Theorem

 3.5.3 [Invertible Matrix Theorem]

For an  matrix , the following are equivalent.

(1) RREF

(2)  is a product of elementary matrices.

(3)  is invertible.

(4)  is the unique solution to .

(5)  has a unique solution for any .

(6) The columns of  are linearly independent.

(7) The rows of  are linearly independent.

 

[Remark]

The vectors of the solution space of  are orthogonal to the rows of .

 

 

 

 

  Let us think of the homogeneous system  with  variables. If the system has  linear equations, then the size of matrix  is . It can be rewritten using inner product. Let  indicate rows of a matrix .

  Thus () if  is a solution to . That is, the vectors in this solution space of  are all orthogonal to the row vectors of the matrix .

 

 

 

 

 



Consider the system of linear equations: 

             ,   .

It is easy to check that  is non-trivial solution of this system. 

Let us verify that  is orthogonal to row vectors of the coefficient matrix  of the above system.  




  0

  0

  0

 Thus  is orthogonal to row vectors of the coefficient matrix .

 

[Remark]

Line, Plane, Hyperplane

 

 

 

 

 (1) Line of -plane: the solution set of a linear equation  where 

 

 (2) Plane of -space: the solution set of a linear equation    where 

 

Note: The solution set of  in  where  forms a plane. If , then it is a hyperplanepassing through the origin. This hyperplane can be considered as a solution set of  for a nonzero vector . This solution set  is called an orthogonal complement of  (or  perp, ), and  is called a normal vector of the hyperplane , which is .

 

 

  : a normal vector of the hyperplane , which is  = the orthogonal complement of .

 

 

 

 

 

 

 

 

 

3.6 Special matrices

 

 Reference video: https://youtu.be/FNRT0d_c9Pg (http://youtu.be/daIxHJBHL_g)

 Practice site: http://matrix.skku.ac.kr/knou-knowls/CLA-Week-4-Sec-3-6.html

 

 

 

 

 

 

We saw various properties of matrix operations. In this section, we introduce special matrices and consider some of their crucial properties.


 Diagonal matrix: A square matrix with the entries 0 except the main diagonal.

 

    A diagonal matrix with its main diagonal entries  can be written as  diag

  

                                    diag

 

 Identity matrix: the matrix with its main diagonal entries all 1’s, denoted by 

 

 Scalar matrix: 

 

                                         ,   

 

 

The following are all diagonal matrices.  and  are scalar matrices.

 

             

 and  are written as  and

.




[ 2  0]          [-3  0  0]

[ 0 -1]          [ 0 -2  0]

                  [ 0  0  1]                                      ■

Consider the following matrix.

If  and ,

 .

 

For a general matrix  is obtained by multiplying each row of  by the corresponding entry of ,

and  is obtained by multiplying each column of  by the corresponding entry of ,

 

Furthermore, it satisfies the following.

 

  

 

 

In other words, the power of a diagonal matrix is the same as the diagonal matrix with the powers of the entries of the main diagonal.                                                     


 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080




D^(-1)=

[   1    0    0]

[   0 -1/3    0]

[   0    0  1/2]

D^5=

[   1    0    0]

[   0 -243    0

[   0    0   32                                                ■


 

Definition

 

 

 

 

 

If a square matrix  satisfies  is called a symmetric matrix. If , then  is called a skew-symmetric matrix.

 

 

 

 

 

In the following matrices,  and  are symmetric matrices and  is a skew-symmetric matrix.

           

● http://matrix.skku.ac.kr/RPG_English/3-SO-Symmetric-M.html 


 http://sage.skku.edu or http://mathlab.knou.ac.kr:8080




 

If  is a square matrix, prove the following.

(1)  is a symmetric matrix.

(2)  is a skew-symmetric matrix.

 (1) Since  is

           a symmetric matrix.

(2)Since is

a skew-symmetric matrix.                                

 

 

 

[Remark]

 

 

 

 

 

A given matrix can be written uniquely as a sum of a symmetric matrix and a skew-symmetric matrix.

 

 

 

 

 

 For any given matrix   and  is a symmetric matrix and

      is a skew-symmetric matrix.                   

 

 Upper triangular matrix: A square matrix whose entries under the main diagonal are all zeros

 Lower triangular matrix: A square matrix whose entries above the main diagonal are all zeros

 

In general,  triangular matrices are as follows.

 

Theorem

 3.6.1 [Property of a triangular matrix]

Let  and  be a lower triangular matrix.

(1)  is a lower triangular matrix.

(2) If  is an invertible matrix, then  is a lower triangular matrix.

(3) If  for all , then the main diagonal entries of  is all 1’s. 


Let  be a square matrix. If there exists an positive integer  such that ( is called nilpotent),  is invertible and . This is because

 

                                   



● http://matrix.skku.ac.kr/LA-Lab/index.htm 

● http://matrix.skku.ac.kr/knou-knowls/cla-sage-reference.htm 



    [Solution]  Section 3-1  http://youtu.be/LaAAruKbGyc

               Section 3-2  http://youtu.be/-MPszmMNvLE

               Section 3-3  http://youtu.be/ceI80eXp6xU

               Section 3-4  http://youtu.be/s7jxVvVAel4

               Section 3-5  http://youtu.be/IygHFdWacds

               Section 3-6  http://youtu.be/rYBsPkeVhQ0

       

          

 

 

  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) I = (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.


 When , confirm the following.        .

 i)  

ii)   

                                                          ■




Sage) http://math1.skku.ac.kr/home/math2013/297/

 

 

 When   , confirm that and that .


         but        


Sage) http://math3.skku.ac.kr/home/pub/20


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]                                                                

 

 

 When , compute the following.

             

  

 

∴ Answer is                                  ■

 

Sage)




                                                                        ■

 


 Show that  is the inverse of . And confirm that .

            

 ====

 

=

=

 

 

 

 = 

==

 

=      ■

 

 

 If , show that .

 

∴ If , then .                                                          ■

 


Solved by 주영은, 김원경, 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

print "E2 ="

print E2

print

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                                  ■



 Using elementary operations, find the inverse of the following matrix.

            (1)      (2)  

(1)  =  → →  ==

(2)  = 

→  → 

.  .              


 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 .               

 

 

 

 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 .                               ■

 

 

 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) .   ■

 


 Give a solution by finding the inverse of the coefficient matrix of the system.

           

 Set the coefficient matrix .

 

   Use ERO to get  :

 

          

 

                                                            

 

                                Ans) =      ■

Sage ) Find Inverse

 

 

       
[ 2/3 -5/3  4/3]
[   0   -1    1]
[   1   -5    4]


        
    



Sage ) Find solution set (해를 구하는 방법)

x= (5/3, 1, 4)

x= (5/3, 1, 4)



 

 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.    ■

 

 

 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) 




[ 1/2    0    0]
[   0 -1/5    0]
[   0    0  1/3]

 


 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    



 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 ?

 

      => 

                                 ■

 

 

 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.  



 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.

 

 

 

 Find  matrices  and explain the relation with ERO.

            

 

  

    ■

 

 

 

 Decide if the following 4 vectors are linearly independent.

            

  

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]                     ■

 


 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.   ■

 

 

 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 .

 

 

 

 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   for all  ().  ■

 

 

 

 If  is invertible and symmetric, so is .

 

     and  .

       =>   =>  ​ =>  is symmetric.       ■

 

 

 


 

                                   Version 2

           Mar. 11, 2016




 

 

 

 About the Author

http://www.researchgate.net/profile/Sang_Gu_Lee

https://scholar.google.com/citations?user=FjOjyHIAAAAJ&hl=en&cstart=0&pagesize=20

http://orcid.org/0000-0002-7408-9648

http://www.scopus.com/authid/detail.uri?authorId=35292447100

http://matrix.skku.ac.kr/sglee/vita/LeeSG.htm