Math: Linear Alegebra Related Question

[surferdude89]

Homework Statement

Hello All,

This question is a problem I ran across and I am working on for practice, but I am having a rough time getting started because of not understanding the context of the problem, Some help would be greatly appreciated in understanding the question.


Suppose $\mathbf{A}$ is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that the *polynomial matrix*

$p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I} $


has the eigenvalues


$p(\lambda_{j}) = k_{m}{\lambda_{j}}^{m}+k_{m-1}{\lambda_{j}}^{m-1}+\ldots+k_{1}\lambda_{j}+k_{0}$


where $j = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$.

Thank You.

Homework Equations

The Attempt at a Solution

If $X$ is an eigenvector of $A$, say $AX=\lambda X$, then we can use that to simplify $p(A)X$ into $(\text{some scalar value})*X$, and that scalar in front of the $X$ is then an eigenvalue of $p(A)$, corresponding to the eigenvector $X$.

This is in LaTeX format of writting. I am not sure this site supports it. Let's see.

 

[lanedance]

i had to write this out to read it, click on the latex to see the code

so A is an nxn matrix, with pontentially non-disctinct eigenvalues \lambda_i

can it by shown that polynomial martix
p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I}

 

[lanedance]

If $X$ is an eigenvector of $A$, say $AX=\lambda X$, then we can use that to simplify $p(A)X$ into $(\text{some scalar value})*X$, and that scalar in front of the $X$ is then an eigenvalue of $p(A)$, corresponding to the eigenvector $X$.

that sounds reasonable to me and shows x is an eignevector of A and p(A),

that only bit that gets me is why non-distinct eigenvalues are mentioned... i think it still works but you need to assume you have a full set of eigenvectors.

though having a non-full set of eignevectors, would mean it has the same eigenvalues for any eigenvector, you may just need it is defective the same as the original matrix, so it doesn't have any other eigenvalues... not too sure on this..

 

[lanedance]

this may help
http://en.wikipedia.org/wiki/Defective_matrix

 

[HallsofIvy]

On this board, use [ itex ] to begin and [ /itex ] to end LaTex (without the spaces)
What you posted was:

Suppose \mathbf{A} is an n \times n matrix with (not necessarily distinct) eigenvalues \lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}. Can it be shown that the *polynomial matrix*

p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I}


has the eigenvalues


p(\lambda_{j}) = k_{m}{\lambda_{j}}^{m}+k_{m-1}{\lambda_{j}}^{m-1}+\ldots+k_{1}\lambda_{j}+k_{0}


where j = 1,2,\ldots,n and the same eigenvectors as \mathbf{A}.

Yes, that's true. It is easy to show, by induction, say, that if v is an eigenvalue of A with eigenvector \lambda, then A^nv= \lambda^n v. It can be shown by direct computation that if v is an eigenvector of both A and B with eigenvalues \lambda_A and \lambda_B, respectively, then v is an eigenvector of pA+ qB with eigenvalues p\lambda_A+ q\lambda B where p and q are any scalars.

I recommend that you try proving those two things yourself.