spectrum of $A-\mu I$

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

Theorem 1.
 $\sigma(A-\mu I)=\{\lambda-\mu\colon\lambda\in\sigma(A)\}$

Theorem 1 is equivalent to:

Theorem 2.

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

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 a spectral value of $(A-\mu I)$, as desired. ∎

Title spectrum of $A-\mu I$ SpectrumOfAmuI 2013-03-22 15:32:49 2013-03-22 15:32:49 PrimeFan (13766) PrimeFan (13766) 9 PrimeFan (13766) Theorem msc 15A18 SpectralValuesClassification