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trying to do some homework but I can't make any sense of this problem.

Suppose that $C$ is an invertible matrix (such that $CAC^{-1}$ is defined). Find an eigenvalue for the matrix $B$ where $B = CAC^{-1}$.

Since its invertible there is no $0$ eigenvalue, $CAC^{-1}$ is defined I don't understand how $B$ could be $CAC^{-1}$.

Suppose that a square matrix 𝐴 has a characteristic polynomial $(\lambda-2)^3(\lambda-4)(\lambda-5)$

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  • $\begingroup$ Are you given that $A$ is a diagonal matrix? $\endgroup$
    – Seeker
    Commented Nov 28, 2022 at 2:34
  • $\begingroup$ I am not given a diagonal matrix no, just a polynomial - updated with polynomial given. $\endgroup$
    – Brandon Zywoo
    Commented Nov 28, 2022 at 2:45
  • $\begingroup$ @BrandonZywoo Is the polynomial the characteristic polynomial of $A$? $\endgroup$
    – Seeker
    Commented Nov 28, 2022 at 2:50
  • $\begingroup$ Yes it is! sorry $\endgroup$
    – Brandon Zywoo
    Commented Nov 28, 2022 at 2:51

1 Answer 1

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First we have to show that $A$ and $B$ share eigenvalues.

Let $v$ be an eigenvector of $A$. Then, $$(CAC^{-1})(Cv)=(CAC^{-1}C)v=(CA)v=C(Av)$$ $$=C(\lambda v)=\lambda Cv$$ Thus, $Cv$ is an eigenvector of $B$ with eigenvalue $\lambda$. Thus, $B$ shares eigenvalues with the matrix $A$.

Now you are given the characteristic polynomial of $A$. The zeros of that polynomial are the eigenvalues of $A$. Since we have showed that all the eigenvalues of $A$ are the eigenvalues of $B$, what can you say about the eigenvalues of $B$?

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