Similar Matrices 

Two matrices A and B are said to be similar if there exists a non-singular matrix P such that  

.     

We all ready know that if a matrix A has distinct eigenvalues then it has a full set of eigenvectors and there exits a matrix P such that where D is a diagonal matrix with eigenvalues on the diagonal and P is a matrix made up of the eigenvectors listed in the same order as the eigenvalues.   Why should we be interested in similar matrices in this particular Chapter or better stated what property do similar matrices share with eigenvalues and eigenvectors.  

We will now assume that A and B are similar matrices and look at how there eigenvalues are related.  The eigenvalues of A are found by looking at the characteristic equation of the matrix A 

i.e.  .  As we know the eigenvalues are the roots of this equation or the roots of the characteristic polynomial.  Note that = = 

.  Now  det() = det() = =  

.  We see that two matrices that are similar must have the same characteristic equation and hence the same set of eigenvalues.  Note nothing has been said about having the same set of eigenvectors.  The eigenvalues are the same but the eigenvalue can be different.   

Our first example will give two matrices A and B along with the matrix P.  We will verify that they are similar and that they have the same characteristic equation and hence eigenvalues. 

Example 1 

>	with(LinearAlgebra):

 

>	A:=<<1,1>|<1,1>>;

 

>	B:=<<-1,1>|<-3,3>>;

 

>	P:=<<1,0>|<2,1>>;

 

>	MatrixInverse(P).A.P;

 

>	P.B.MatrixInverse(P);

 

>	I2:=IdentityMatrix(2);

 

>	p:=Determinant(A-lambda*I2);

 

>	pa:=Determinant(B-lambda*I2);

 

>	lambda[1]:=0;

 

>	lambda[2]:=2;

 

>	Eigenvalues(A);

 

The eigenvalues are given as a vector with each entry a eigenvalue.  We can use the following command to get the eigenvalues reported as a list. 

>	Eigenvalues(A,output='list');

 

(2.1)

 

We would now like to know if two matrices that have the same eigenvalues can we find a matrix P that will make the two similar.   

Finding the matrix P 

Next we investigate how to find the matrix P above.  The answer lie in being able to make A and B similar to same diagonal matrix.   

First we need to find the eigenvector of the matrix A 

>	A1:=A-0*I2;

 

>	A1a:=NullSpace(A1);

 

>	p1:=A1a[1];

 

So we see that a = -b which say that an eigenvector is [1,-1].  The eigenvector is a basis for the nullspace of the matrix A1.   

>	A2:=A-2*I2;

 

>	A2a:=NullSpace(A2);

 

>	p2:=A2a[1];

 

(4.1)

 

This means that a second  eigenvector is [1,1].  Again the eigenvector is a basis vector for the nullspace of the matrix A2.   

>	Eigenvectors(A);

 

As you can see the Eigenvectors command first gives the eigenvalues as a vector and then gives the eigenvector as the first column of a matrix.  The second value in the column vector is the second eigenvalue and the second column of the matrix is an eigenvalue associated with 2.  Here is anoter version of the command. 

>	Eigenvectors(A,output='list');

 

(4.2)

 

There is a problem with this command also.  We do not know what the 1 in each of the groups mean at this time.  The first entry is always the eigenvalue and the vectors are associated eigenvectors for that eigenvalue.  We will be at a point shortly to explain what the 1 means in each of the groups.  For not we will not worry about it.   

>	P1:=<<1,-1>|<1,1>>;

 

>	MatrixInverse(P1).A.P1;

 

The above process will now be carried out for the matrix B starting with the eigenvalue 0.  

>	B1:=B-0*I2;

 

>	B12:=NullSpace(B1);

 

>	p21:=B12[1];

 

(4.3)

 

An eigenvector will be [-3,1].  As above we now look for an eigenvector associated with the eigenvalue 2.  

>	B2:=B-2*I2;

 

>	B2a:=NullSpace(B2);

 

>	p22:=B2a[1];

 

(4.4)

 

An eigenvector will be [1,-1]. 

Now we will look at the Maple command 

>	EB:=Eigenvectors(B);

 

(4.5)

 

Here is a way to get the P2 matrix from EB 

>	P2:=EB[2];

 

>	MatrixInverse(P2).B.P2;

 

This tell us that = .  The diagonal matrix D must be the same for both the diagonalization of A and the diagonlaization of B.   We now see that
Multiplying on the left by and on the right by we obtain 

.    Set and observe that (why?)    Recall that  

We now return to our example. 

>	P:=P2.MatrixInverse(P1);

 

>	MatrixInverse(P).B.P;

 

>	Pa:=P1.MatrixInverse(P2);

 

This appears to be different from the very first P that we found.  Does it make a difference?  

. 

>	MatrixInverse(Pa).A.Pa;

 

A word of warning we can not always accomplish the above.  We will need to see when this can be done.  

Consider the following two matrices .  

>	Da:=DiagonalMatrix(<2,2>);

 

>	B:=<<2,0>|<1,2>>;

 

A quick inspection tell us that both matrices have only one eigenvalue of 2   

What are the vectors associate with the eigenvalue of 2 for the matrix Da 

We see that we get the zero matrix when we subtract off 2 from the diagonal.  Do not panic 

This tells us that any two linearly independent vector will work for eigenvectors i.e. 

[1,0],[0,1] and we have the identity matrix for P. 

Now let us look for eigenvectors when when look at matrix B 

>	B1:=B-2*I2;

 

>	NullSpace(B1);

 

(4.6)

 

The eigenvector associated with 2 will be all of the from [1,0].  There is only one so it can not be made similar to the diagonal matrix Da and hence they are not similar .  This will lead to our next topic but first we will look at an example for a 3x3 matrix. . 

Gersgorin Circle Theorem and Applications 

Theorem (Gersgorin Circle) Let A be an matrix and denote the circle in the complex pland with center and radius   ||:  that is  

                        = {z in C | || < = || } 

where C denotes the complex plane.  The eigenvalues of A are contained within R =   .  Moreover, 

the union of any of these circles that do not interest the remaining () contains precisely (counting multiplicities) of the  eigenvalues.   

>	with(LinearAlgebra):

 

>	A:=<<-4,0,-2>|<1,2,0>|<1,1,9>>;

 

We see from the above theorem that there are three circle to consider.  

1.  A circle centered  at -4 with radius 2 

2.  A circle centered  at 2 with radius 1 

3.  A circle centered at 9 with radius 2 

Look at the graph.   

There are three real eigenvalues.  Why?  The matrix can be diagonalized.  Why? 

>	evalf(Eigenvalues(A));

 

The above illustrates that you need to know a some theory and not trust a machine all the time. 

Knowing something about the matrices also makes the software very useful.    The eigenvalues are all real.  The 10 to the negative power is just round off error and the real part is the true eigenvalues.   

>	AA:=<<4,0,-2>|<1,2,0>|<1,1,9>>;

 

There are now two circles that interect and one that stands alone. 

1.  A circle centered at 4 with radius 2  

2.  A circle centered at 2 with radius 1 

3   A circle centered at 9 with radius 2 

The circles centered at 4 and 2 interect so the best we can say is that there are two eigenvectors in the union of the two circles.  We could have complex eigenvalues in this case.   There is one real eigenvalue in the circle centered about 9.   

>	evalf(Eigenvalues(AA));

 

>	AAA:=<<9,0,0>|<0,0,3>|<1,-3,0>>;

 

The circle for this matrix are as follows 

1.  A circle centered at 9 with radius 1  

2.  A circle centered at 0 with radius 3.   

>	evalf(Eigenvalues(AAA));

 

>	AAAA:=<<9,0,0>|<0,0,3>|<1,3,0>>;

 

>	Eigenvectors(AAAA);

 

>

 

The two of the eigenvalues of AAA are complex and appear in complex conjugate pairs. i.e.  a+ib , a-ib.  here 3i and -3i. The eigenvalues of AAAA are real i.e. -3,9,3.  Again an eigenvector of each eigenvalue is given in the matrix for AAAA.   

Exercises 3.2 

Exercise 3.2.1 , ( Instead of Exercise 3.2.3 on page 367) Do the following two problmes by  following  the instructions for Exercise 3.2.3 

a. b. , Exercise 3.2.4, 3.2.11, 3.2.13  Exercises are due at the end of class on November 12, 2008,  You can get them to me by e-mail.  

>