spectrum of $A-\mu I$

Let $A$ be an endomorphism of the vector space$V$ over a field $k$. Denote by $\sigma (A)$ the spectrumof $A$. Then we have:

Theorem 1.

$$\sigma (A-\mu I)=\{\lambda -\mu :\lambda \in \sigma (A)\}$$

Theorem 1 is equivalent to:

Theorem 2.

$\lambda$ is a spectral value of $A$ if and only if$\lambda \mathrm{-}\mu$ is a spectral value of $A\mathrm{-}\mu \mathit{}I$.

Proof of Theorem 2.

Note that

$$A-\lambda I=(A-\mu I)-(\lambda I-\mu I)=(A-\mu I)-(\lambda -\mu )I$$

and thus $A-\lambda I$ is invertible if and only if$(A-\mu I)-(\lambda -\mu )I$ is invertible. Equivalently,$\lambda$ is a spectral value of $A$ iff $\lambda -\mu$ is aspectral value of $(A-\mu I)$, as desired.∎