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Homespectrum of $A-\mu I$

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# 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. ∎

Related:

SpectralValuesClassification

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

## Mathematics Subject Classification

15A18*no label found*

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