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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
spectrum of 
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(Theorem)
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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:
Proof. [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. 
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"spectrum of  " is owned by PrimeFan. [ full author list (5) | owner history (2) ]
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Cross-references: iff, invertible, spectral value, equivalent, theorem, spectrum, field, vector space, endomorphism
This is version 6 of spectrum of  , born on 2005-10-22, modified 2007-02-11.
Object id is 7444, canonical name is SpectrumOfAMuI.
Accessed 1941 times total.
Classification:
| AMS MSC: | 15A18 (Linear and multilinear algebra; matrix theory :: Eigenvalues, singular values, and eigenvectors) |
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Pending Errata and Addenda
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