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Say we have a matrix $\mathbf{B}$

$$ \mathbf{B}=\begin{pmatrix} 6&0&0\\ 2&-3&-4\\ -5&2&3 \end{pmatrix} $$

We know the eigenvalues are $\lambda_1=6$, $\lambda_2=-1$ and $\lambda_3=1$. We're then asked to express the eigenvalues for

$$2\mathbf{B}^2+\mathbf{I}$$

I assume that for any matrix raised to the power $k$ (where $k$ is a positive integer), it's associated eigenvalues are $\lambda^k$. I also assume that for any matrix multiplied by some number $k$, it's associated eigenvalues will be $k\lambda$. Finally, for any matrix added to a multiple (which is any number for $k$) of an identity matrix, $\mathbf{B}+k\mathbf{I}$, the eigenvalues will be $\lambda+k$. So for the eigenvalue $\lambda_1=6$ for $2\mathbf{B}^2+\mathbf{I}$ is

$$ 2\times(6)^2+1=73 $$

Have I correctly worked this out?

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  • $\begingroup$ only if you know how to prove all your 'assume' statements $\endgroup$
    – Exodd
    Dec 4, 2018 at 12:35
  • $\begingroup$ link This might help. More general and has a way to prove $\endgroup$
    – mmcrjx
    Dec 4, 2018 at 12:36
  • $\begingroup$ Do you know the Jordan-form? $\endgroup$ Dec 4, 2018 at 12:47
  • $\begingroup$ @whitelined the title of your question doesn't really express what you are looking for. Could you reword it e.g. Use known eigenvalues of $B$ to express those of $B^2+I$ $\endgroup$
    – user376343
    Dec 4, 2018 at 13:04

2 Answers 2

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You're right. If $(\lambda,u)$ is an eigenpair of $\bf{B},$ then $$\begin{aligned}\left(\bf{B}^2+\bf{I}\right)u&={\bf{B}}({\bf{B}}u)+{\bf{I}}u\\ &={\bf{B}}\lambda u+u\\&=\lambda({\bf{B}}u)+u\\&=(\lambda ^2+1)u\end{aligned}$$

Thus $(\lambda ^2 +1, u)$ is an eigenpair of $\bf{B}^2+\bf{I}.$

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You are right. A generalization reads as follows: If $A$ is a real or complex square matrix then we denote the set of eigenvalues of $A$ by $\sigma(A)$. If $p$ is a polynomial then we have the "spectral mapping theorem":

$$ \sigma (p(A)) =p( \sigma(A)).$$

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