# 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.

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

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

Title | spectrum of $A-\mu I$ |
---|---|

Canonical name | SpectrumOfAmuI |

Date of creation | 2013-03-22 15:32:49 |

Last modified on | 2013-03-22 15:32:49 |

Owner | PrimeFan (13766) |

Last modified by | PrimeFan (13766) |

Numerical id | 9 |

Author | PrimeFan (13766) |

Entry type | Theorem |

Classification | msc 15A18 |

Related topic | SpectralValuesClassification |