# spectral mapping theorem

Let $\mathcal{A}$ be a unital ${C}^{*}$-algebra (http://planetmath.org/CAlgebra). Let $x$ be a normal element in $\mathcal{A}$ and $\sigma (x)$ be its spectrum.

The continuous functional calculus provides a ${C}^{*}$-isomorphism^{}

$C(\sigma (x))\u27f6\mathcal{A}[x]$

$f\mapsto f(x)$

between the ${C}^{*}$-algebra $C(\sigma (x))$ of complex valued continuous functions^{} on $\sigma (x)$ and the ${C}^{*}$-subalgebra $\mathcal{A}[x]\subseteq \mathcal{A}$ generated by $x$ and the identity of $\mathcal{A}$.

Spectral Mapping Theorem - Let $x\in \mathcal{A}$ be as above. Let $f\in C(\sigma (x))$. Then

$$\sigma (f(x))=f(\sigma (x)).$$ |

Proof : Since $C(\sigma (x))$ and $\mathcal{A}[x]$ are isomorphic we must have

$$\sigma (f)={\sigma}_{\mathcal{A}[x]}(f(x))$$ |

where ${\sigma}_{\mathcal{A}[x]}(f(x))$ denotes the spectrum of $f(x)$ relative to the subalgebra $\mathcal{A}[x]$.

By the spectral invariance theorem we have ${\sigma}_{\mathcal{A}[x]}(f(x))=\sigma (f(x))$. Hence

$$\sigma (f)=\sigma (f(x))$$ |

Thus, we only have to prove that $f(\sigma (x))=\sigma (f)$.

$f$ is defined on $\sigma (x)$ so $f(\sigma (x))$ is precisely the image of $f$.

Let $\lambda \in \u2102$. The function $f-\lambda $ is invertible if and only if $f-\lambda $ has no zeros.

Equivalently, $f-\lambda $ is not invertible if and only if $f-\lambda $ has a zero, i.e. $f({\lambda}_{0})=\lambda $ for some ${\lambda}_{0}$.

The previous statement can be reformulated as: $\lambda \in \sigma (f)$ if and only if $\lambda $ is in the image of $f$.

We conclude that $\sigma (f)=f(\sigma (x))$, and this proves the theorem. $\mathrm{\square}$

Title | spectral mapping theorem |
---|---|

Canonical name | SpectralMappingTheorem |

Date of creation | 2013-03-22 17:30:08 |

Last modified on | 2013-03-22 17:30:08 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 4 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46L05 |

Classification | msc 47A60 |

Classification | msc 46H30 |