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Let $A \in \text{M}_{n \times n}(\mathbb{C})$. How do I see that if $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $A$, then for any polynomial $P(\cdot)$, I have that $P(\lambda_1), \dots, P(\lambda_n)$ are the eigenvalues for $P(A)$?

Eugene Zhang
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    let $v$ be an eigenvector to eigenvalue $\lambda$. Then, $P(A)v = \dotsb = P(\lambda)v$ – user251257 Sep 30 '15 at 02:00
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    http://math.stackexchange.com/questions/492070/how-to-prove-eigenvalues-of-polynomial-of-matrix-a-polynomial-of-eigenvalue?rq=1 – daw Sep 30 '15 at 07:27

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