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If I know the characteristic polynomial of a matrix $A$, what can I know about the charpoly of $A^2$? And if I have the charpolys of $A$ and $B$, what can I know about the charpoly of $A+B$? I'm trying to solve the following problem:

The eigenvalues of $A$ are $1,-3,0$. Show that the eigenvalues of $A^2+A-2I$ are $0,2,-4$.

Thank you!

Edit: I now know that the eigenvalues of $A^2$ are the squares of the eigenvalues of $A$. I still need help solving the problem. Thanks!

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3 Answers

If the eigenvalues of $A$ are $c_1, \ldots, c_n$ then the eigenvalues of $A^k$ are $c_1^k, \ldots, c_n^k$. You can see this putting $A$ in Jordan form and using the fact that the diagonal entries of a power of a triangular matrix T are the powers of the diagonal entries of that matrix T.

More generally, if $p$ is a polynomial then the eigenvalues of $p(A)$ are $p(c_1), \ldots, p(c_n)$ by the same reasoning above.

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Thanks! and what can I say about the eigenvalues of A+B? of A+I? Is it the sums of the eigenvalues of the matrices? –  idan Jun 16 '12 at 7:20
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Given a matrix $A$ with eigenvalue $\lambda$, and any polynomial $p(X) = \sum_{k=0}^n a_k X^k$, you can show that $p(\lambda)$ is an eigenvalue of $p(A) = \sum_{k=0}^n a_k A^k$.

Just take an eigenvector $v \ne 0$ of $A$ and look at what $p(A) v$ is.

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Thanks! and what can I say about the eigenvalues of A+B? of A+I? Are they the sums of the eigenvalues of the matrices? How can I prove it? –  idan Jun 16 '12 at 7:26
    
$A + I$ would be an example of a polynomial $p(A)$. –  Cocopuffs Jun 16 '12 at 7:27
    
In general, there's not a whole lot you can say about the eigenvalues of $A+B$ other than some numerical bounds in certain cases –  Cocopuffs Jun 16 '12 at 7:28
    
I understand. Then, in my exercise, what can I say about the eigenvalues of A^2+A-2I? I noticed that it's equal to (A+2I)(A-I), does that help? –  idan Jun 16 '12 at 7:32
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For the eigenvalues: If $Ax= \lambda x$, then $ A^2x = A\lambda x = \lambda^2x $. Does that help you?

Edit: I see that you still have problems, so I will find one of the eigenvalues. Let $x$ be the eigenvector correpsonding to the eigenvalue 1. Then $$ (A^2 + A - 2I)x = A^2x + Ax - 2Ix = 1^2x + 1x - 2x = (1+1-2)x = 0x, $$ So $0$ is an eigenvalue of $A^2 + A - 2I$. The other eigenvalues are -2 and 4. Can you prove that?

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Thanks! and what can I say about the eigenvalues of A+B? of A+I? Are they the sums of the eigenvalues of the matrices? How can I prove it? –  idan Jun 16 '12 at 7:26
    
Use another definition of eigenvalues, and find an eigenvector. –  awllower Jun 16 '12 at 7:30
    
If $x$ is an eigenvector for both $A$ and $B$ with eigenvalues $\lambda_A$ and $\lambda_B$, then $(A+B)x = Ax + Bx = (\lambda_A + \lambda_B)x$, and the eigenvalue for $A+B$ is $(\lambda_A + \lambda_B)$ with eigenvector $x$. IF they have different eigenvectors, I don't know if there is a trivial way to find the eigenvalues of $A+B$. –  N.U. Jun 16 '12 at 7:30
    
and about multiplication of matrices? –  idan Jun 16 '12 at 7:40
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