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Let $A$ be a square matrix and $p(\lambda)$ any polynomial. If $\lambda$ is an eigenvalue of $A$, show that $p(\lambda)$ is an eigenvalue of the matrix $p(A)$ with the same eigenvector.
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Suppose that $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, i.e. $Av = \lambda v$.

Then, $p(A)v = p(A) \cdot v = p(\lambda)v$, so $p(\lambda)$ is an eigenvalue of $p(A)$ with eigenvector $v$.
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