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MathsGee Android Q&A


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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.
in Mathematics by Platinum (164,236 points) | 301 views

1 Answer

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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 $v$.
ago by Platinum (164,236 points)

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MathsGee Android Q&A

MathsGee Android Q&A