The eigenvalues of $A+I$

Question

I recently read the following claim:

Let $A \in \mathbb{R}^{n \times n}$. If $\mu$ is an eigenvalue of $A + I$ then $(\mu - 1)$ is an eigenvalue of $A $.

I do not see why this should be true. Is this true at all?

Answer 1

There is an obvious typo in the question. At the end you have typed $A+I$ instead of $A$.

There exists a non-zero vector $x$ such that $(Ax+I)=\mu x$. Hence $Ax=(\mu- 1)x$, so the answer is YES.