Algebraic and Geometric Multiplicity of Eigenvalues 

We want to take a closer look at when we can diagonalize a matrix.   

Algebraic Multiplicity of Eigenvalues 

The algebraic multiplicity of an eigenvalue is the power of the eigevalue factor in the characteristic polynomial.   

Geometric Multiplicity of an Eigenvalue 

The Geoemtric Multiplicity of an eigenvalue is the number of the basis vectors  in  the nullspace for the matrix for a given eigenvalue .   

A matrix can be made diagonally similar to a diagonal matrix with eigenvalue on the diagonal if and only if the algebraic and geomemtric multiplicity of an eigenvalue are equal. 

This tell us that if the eigenvalues are all of algebraic multiplicity one then the matrix is always similar to a diagonal matix with the eigenvalues one the diagonal.  We will illustrate this  fact with several illustrations.   

Example 1 

This first example will be a symmetric matrix.  We will look for an orthogonal matrix instead of a matrix P i.e. We want to find Q such that .  When the matrix is symmetric you can always fid an orthogonal matrix.  When the matrix is not symmetric then it will not be able to find a Q but you will be able to find a P if the algebraic multiplicity equal the geometric multiplicity.   Recall that an orthogonal matrix Q is a matrix where the columns form a orthogonal set of vectors and each as lenght one.  That is the vectors are orthogonal and they have been normalized.   

>	with(LinearAlgebra):

 

>	A:=<<2,0,0>|<0,0,2>|<0,2,0>>;

 

>	I3:=IdentityMatrix(3);

 

>	Alambda:=A-lambda*I3;

 

>	Determinant(Alambda);

 

>	factor(%);

 

Algebraic multiplicity of 2 is 2 while the algebraic multiplicity of -2 is one. 

>	A2:=subs(lambda=2,Alambda);

 

We will look at a long way to find the eigenvector(s) for the eigenvalue -2.  First you should understand that the first column has all zeros which tell you right away that the general solution in the nullspace is arbitary.  If we use matrix B below then a is arbitrary.  Notice that the second and third row only differ by a negative sign so they are not different.  We will multiply the matrix A by the vector B and take a closer look.   

>	B:=<a,b,c>;

 

>

A2.B;

 

We have something happen that is very very important.  The 0 in the above vector tell us that a is arbitrary.   The next two entries are reall the same which is that 2b-2c=0 or b=c. 

The vectors in the null space thus look like  

=a + b .  This tells us that we have two basis vectors in the null space. 

The geometric multiplicity of is 2 i.e. the geometric multiplicity and algebraic multipliciy are the same.   

We need now to look at the eigenvaul .  Notice every eigenvalue must have at least one eigenvector.   We know that this eigenvector is linearly independent to the two from the other eigenvalue.  I took the longer way so you would get a better feel for what was happening.  You should look at some of the exercises in this longer way to get a better feel for what is going on.  We can aways use the NullSpace command.   

>	NullSpace(A2);

 

(2.1)

 

We get the same two vectors as in the longer method.   

>	A3:=subs(lambda=-2,Alambda);

 

With the 4 in the (1,1,) position we see that the first entry in the nullspace of this matrix must be zero and the next two line tells us that the next two entries in the eigenvector must be equal and opposite in sign.  We will us the following eigenvector.  .  Instead of using  

P= we will normailize the columns of P and use the orthogonal matrix.  

Q= .   Take a close look at why we choose to use this representation of the matrix.   

>	NullSpace(A3);

 

(2.2)

 

>	Q:=<<1,0,0>|<0,sqrt(2)/2,sqrt(2)/2>|<0,sqrt(2)/2,-sqrt(2)/2>>;

 

>

Q.A.Q;

 

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

 

The Eigenvectors command with output='list' gives you all the information you need  first it tells you -2 is an eigenvector then the 1 following -2 is the algebraic multiplicity.  Next you get a generator for the eigenspace.  Since in this group there is only one vector it tell you the geometric multiplicity is 1.  In the next group the first element is 2 so you have an eigenvalue of 2.  The next number is again 2 so you that the eigenvalue has algebraic multiplicitly of 2.  Finally there are two eigenvectors listed so you have geometric multiplicity of two.  This means as above that since the algebraic multiplicity of each eigenvalue equals the geometric multiplicity then the matix can be made similar to diagonal matrix with the eigenvalues on the diagonal as we did abve.  the reason I did not take the transpose of the matrix Q is due to the fact that Q is symmetric.   

First entry is the eigenvalue, the second is the algebrais multiplicity and the last is a basis for the nullspace for the matrix for the given eigenvalue .  You also, quickly see the geometric multiplicty by looking at the entries.  We will go over how to extract the vectors from the above equation 

>	p1:=eA[1][3][1];

 

>	p2:=eA[2][3][1];

 

>	p3:=eA[2][3][2];

 

>	Q:=<p2|Normalize(p3,2)|Normalize(p1,2)>;

 

In eA[1][3][1] says to look at the first set of nmbers [1], [3] says look at the third set of numbers in the first set.  Since this set has {[0,-1,1]} it represents a set so you want the first entry.  Even though there is only one entry you must include it to extract the desired vector [0,-1,1].  You should not be able to figure out the way of extracting the two vectors in the second set of entries.   

Example 2  Not diagonalizable. 

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

 

>	AAL:=AA-lambda*I3;

 

>	p:=Determinant(AAL);

 

>	factor(p);

 

>	AAL2:=subs(lambda=2,AAL);

 

It is very important to remember that the zero vector is never never an eigenvector so what is the null space of the matrix AAL2 above.   

The first equation tells you that the last entry must be zero.  The next two equations tell you that the second entry must be equatl to zero.  Does this mean that we just lied to you and the only eigenvector is the zero vector.  Think for just a minute.  What do we know about the first entry?  It is arbitrary.  This tells us that a bais for the nullspace is .   

We see that the algebraic multiplicity was two and the geometric multiplicity is only one.  This tell use that we have a problem trying to make the matrix AA similar to a diagonal matrix and in 

fact it can not be made similar to a diagonal matrix.  Let us look at the other eigenvalue 

>	n1:=NullSpace(AAL2);

 

(3.1)

 

>	AALm2:=subs(lambda=-2,AAL);

 

>	n2:=NullSpace(AALm2);

 

We will multiply n2 by 4 just to obtain whole numbers.  

>	P:=<n1[1]|4*n2[1]>;

 

The matrix is not square so we are not able to find the inverse.  We can not diagonalize this matrix.  We will find that there is something interesting that we can do but that will have to wait for a few days.   

Example 3 

We will look at a matrix that has multiple roots that is not symmetric and where the algebraic multiplicity and geometric multiplicity are equal.  I am going to work out this example just like you would have to do it when you use paper and pencil.  I will even expand out the determinant by using the Laplace expansion of the second row.   

>	B:=<<2,-1,2>|<-2,3,-4>|<-1,1,-1>>;

 

>	BL:=B-lambda*I3;

 

>	p1:=(2-lambda)*Determinant(SubMatrix(BL,2..3,2..3));

 

>	p2:=-2*Determinant(<<-1,2>|<1,-1-lambda>>);

 

>	p3:=-1*Determinant(SubMatrix(BL,2..3,1..2));

 

>	p:=p1-p2+p3;

 

>	factor(p);

 

>	lambda[1]:=2;

 

>	lambda[2]:=1;

 

>	BL1:=subs(lambda=2,BL);

 

>

 

>	BL1aa:=ReducedRowEchelonForm(BL1);

 

Using back substitution you find the we obtain the following vector if we multiply by 2.  

>	p3:=<1,-1,2>;

 

>	BL2:=subs(lambda=1,BL);

 

>	GaussianElimination(BL2);

 

We see that a=2b + c  so the vector in the null space of this matrix or the basis for the eigenvector is given by  b +c.  We see that has algebraic multipliciy 2 and geometric multiplicity 2 .  We have a complelte set of eigenvectors so the matrix B is diagonally similar to a diagonal matrix with the eigenvalues of the diagonal. 

We will give the shortcut command in Maple to check our work.   

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

 

(4.1)

 

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

 

You need to use a matrix P not an orthogonal matrix since we do not have a symmetric matrix.  Never try to get an orthogonal matrix if you do not have a symmetric matrix.  

>	PIN:=MatrixInverse(P);

 

>

PIN.B.P;

 

>

 

Comments About Special Matrices.  

>	C:=<<1,0,0>|<1/2,1/2,0>|<-3,2,-4>>;

 

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

 

As you can see the eigenvalues are on the diagonal.  Since there as all distince you will have algebraic multiplicity and geometric multiplicity of one for each eigenvalue. 

>	CL:=<<2/3,1/2,3/10>|<0,-3/2,4/3>|<0,0,2/3>>;

 

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

 

Again the eigenvalues on on the diagonal for the lower triangular matrix just as they were for the upper triangular matrix.   

>	DD:=DiagonalMatrix(<Pi,exp(1),1/2>);

 

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

 

>

 

>

 

>

 

>

 

Exercises Section 3.3 

Exercise 3.3.1 (This problem need to be done using text.  Trying to find an example means you are missing the point of the problem).  Exercise 3.3.3, Exercise 3.3.5 (a) (b),(d),(g).  Find the eigenvalues by looking at det(A-λI)=0 and solving this equation.  You can use the solve.  Also, find the eigenvectors for a given eigenvalue by finding the nullspace(A-λI) for each eigenvalue.  Use the eigenvects command to check your answer.  3.3.6(a),(e) You can use the eigenvects command on this problem.  Exercise 3.3.7.   

You can use any of the Maple commands but follow the directions of the problem.  Problems are due on November 14, 2008.