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Let $T: \mathbb C^n \rightarrow \mathbb C^n$ be linear. Let $\beta$ and $\gamma$ be any two ordered bases. Prove that the eigenvalues of $[T]_\beta$ and $[T]_\gamma$ are the same.

Can anyone provide tips/hints in the right direction? I'm struggling as I try to understand this intuitively....thank you

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    $\begingroup$ what do you know about similar matrices? $\endgroup$ – user87543 Dec 5 '13 at 5:13
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Hint:

$[T]_\beta$ and $[T]_\gamma$ are similar matrices, i.e. there is a matrix $Q$ (change of coordinate matrix) so that $[T]_\beta = Q^{-1}[T]_\gamma Q$.

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  • $\begingroup$ Oh okay, so they have the same characteristic polynomials and therefore same eigenvalues. Thanks! $\endgroup$ – A A Dec 5 '13 at 5:21
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The eigenvalues are a property of the operator $T$. Recall that $\lambda$ is an eigenvalue of $[T] $ if $0=\det([T]-\lambda\,I)=\det([T-\lambda\,I]) $. This happens precisely when the operator $T-\lambda\, I$ is not invertible, i.e. there exists $v\in\mathbb C$ with $Tv=\lambda v$. This shows that the eigenvalues of $[T]$ (in any basis) are precisely the eigenvalues of $T$.

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HINT $$det(B - \lambda I_n) = det(A - \lambda I_n)$$ for two similar matrices $A$ and $B$. (Why?)

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