# Use known eigenvalues to express eigenvalues of multiple and exponent of matrix

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?

• only if you know how to prove all your 'assume' statements – Exodd Dec 4 '18 at 12:35
• link This might help. More general and has a way to prove – mm-crj Dec 4 '18 at 12:36
• Do you know the Jordan-form? – Peter Melech Dec 4 '18 at 12:47
• @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 $B^2+I$ – user376343 Dec 4 '18 at 13:04

## 2 Answers

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 $$\bf{B}^2+\bf{I}.$$

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":

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