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Let $A: \mathbb C^4 \to \mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?

Please, explain why this is true or false.

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marked as duplicate by Arnaud D., Mostafa Ayaz, Namaste linear-algebra Dec 14 '18 at 20:21

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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – José Carlos Santos Nov 26 '18 at 19:45
  • $\begingroup$ What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$? $\endgroup$ – SZN Nov 26 '18 at 19:47
  • $\begingroup$ Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$. $\endgroup$ – Connor Harris Nov 26 '18 at 20:03
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The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)\cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product $$ (A - z_1 I) \cdots (A - z_d I) $$ fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.

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