nyuuzyou/wikis huggingface dataset:
Spectral Radius is Multiplicative on Commuting Elements in Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x, y \in A$ be commuting elements of $A$.
Let $r_A$ denote spectral radius in $A$.
Then we have:
:$\map {r_A} {x y} \le \map {r_A} x \map {r_A} y$ 
== Proof ==
First take $\struct {A, \norm {\, \cdot \,} }$ to be unital. 
From Existence of Maximal Commutative Subalgebra, there exists a maximal commutative subalgebra $B$ containing $x$ and $y$. 
From Maximal Subalgebra in Normed Algebra is Closed, $B$ is closed.
From Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra we have $\map {\sigma_B} x = \map {\sigma_A} x$.
Hence by the definition of spectral radius, we have $\map {r_B} x = \map {r_A} x$. 
Let $\Phi_B$ be the spectrum of $B$.
Then we have, from Spectral Radius of Element of Commutative Banach Algebra:
:$\ds \map {r_B} {x y} = \sup_{\phi \in \Phi_B} \cmod {\map \phi {x y} }$
Since $\phi$ is a character we have:
:$\map \phi {x + y} = \map \phi x \map \phi y$ 
Hence by the Triangle Inequality for Complex Numbers, we have:
:$\ds \cmod {\map \phi {x y} } \le \cmod {\map \phi x} \cmod {\map \phi y} \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Taking supremums, we have:
:$\ds \sup_{\phi \in \Phi_B} \cmod {\map \phi {x y} } \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Hence:
:$\map {r_B} {x y} \le \map {r_B} x \map {r_B} y$ 
from Spectral Radius of Element of Commutative Banach Algebra.
Since again $\map {\sigma_B} x = \map {\sigma_A} x$ we deduce:
:$\map {r_A} {x y} \le \map {r_A} x \map {r_A} y$ 
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not necessarily unital. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra, $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital Banach algebra.
We then have:
:$\map {r_{A_+} } {\tuple {x y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
From the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
:$\map {\sigma_A} y = \map {\sigma_{A_+} } {\tuple {y, 0} }$
:$\map {\sigma_A} {x y} = \map {\sigma_{A_+} } {\tuple {x y, 0} }$
Hence:
:$\map {r_{A_+} } {\tuple {x y, 0} } = \map {r_A} {x y}$
:$\map {r_{A_+} } {\tuple {x, 0} } = \map {r_A} x$
:$\map {r_{A_+} } {\tuple {y, 0} } = \map {r_A} y$
Hence we obtain:
:$\map {r_{A_+} } {\tuple {x y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
for $\struct {A, \norm {\, \cdot \,} }$ not necessarily unital.
Category:Spectral Radius
Category:Banach Algebras
Spectral Radius is Subadditive on Commuting Elements in Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x, y \in A$ be commuting elements of $A$.
Let $r_A$ denote spectral radius in $A$.
Then we have:
:$\map {r_A} {x + y} \le \map {r_A} x + \map {r_A} y$ 
== Proof ==
First take $\struct {A, \norm {\, \cdot \,} }$ to be unital. 
From Existence of Maximal Commutative Subalgebra: Unital, there exists a maximal commutative unital subalgebra $B$ containing $x$ and $y$. 
From Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra we have $\map {\sigma_B} x = \map {\sigma_A} x$.
Hence by the definition of spectral radius, we have $\map {r_B} x = \map {r_A} x$. 
Let $\Phi_B$ be the spectrum of $B$.
Then we have, from Spectral Radius of Element of Commutative Banach Algebra:
:$\ds \map {r_B} {x + y} = \sup_{\phi \in \Phi_B} \cmod {\map \phi {x + y} }$
By linearity we have:
:$\map \phi {x + y} = \map \phi x + \map \phi y$ 
Hence by the Triangle Inequality for Complex Numbers, we have:
:$\ds \cmod {\map \phi {x + y} } \le \cmod {\map \phi x} + \cmod {\map \phi y} \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} + \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Taking supremums, we have:
:$\ds \sup_{\phi \in \Phi_B} \cmod {\map \phi {x + y} } \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} + \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Hence:
:$\map {r_B} {x + y} \le \map {r_B} x + \map {r_B} y$ 
from Spectral Radius of Element of Commutative Banach Algebra.
Since again $\map {\sigma_B} x = \map {\sigma_A} x$ we deduce:
:$\map {r_A} {x + y} \le \map {r_A} x + \map {r_A} y$ 
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not necessarily unital. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra, $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital Banach algebra.
We then have:
:$\map {r_{A_+} } {\tuple {x + y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
From the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
:$\map {\sigma_A} y = \map {\sigma_{A_+} } {\tuple {y, 0} }$
:$\map {\sigma_A} {x + y} = \map {\sigma_{A_+} } {\tuple {x + y, 0} }$
Hence:
:$\map {r_{A_+} } {\tuple {x + y, 0} } = \map {r_A} {x + y}$
:$\map {r_{A_+} } {\tuple {x, 0} } = \map {r_A} x$
:$\map {r_{A_+} } {\tuple {y, 0} } = \map {r_A} y$
Hence we obtain:
:$\map {r_{A_+} } {\tuple {x + y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
for $\struct {A, \norm {\, \cdot \,} }$ not necessarily unital.
Category:Spectral Radius
Category:Banach Algebras
Spectral Radius of Element of Commutative Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$. 
Let $\Phi_A$ be the spectrum of $A$.
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$. 
Let $\map {r_A} x$ be the spectral radius of $x$.
Then:
:$\ds \map {r_A} x = \sup_{\phi \in \Phi_A} \cmod {\map \phi x}$
== Proof ==
From the definition of the spectral radius, we have:
:$\ds \map {r_A} x = \sup_{\lambda \in \map {\sigma_A} x} \cmod \lambda$ 
From Spectrum of Element of Unital Commutative Banach Algebra, we have:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$ if $A$ is unital
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$ if $A$ is not unital.
Hence we have:
:$\ds \map {r_A} x = \max \set {\sup_{\phi \in \Phi_A} \cmod {\map \phi x}, 0} = \sup_{\phi \in \Phi_A} \cmod {\map \phi x}$
Category:Banach Algebras
Category:Spectral Radius
Spectral Radius of Normal Element of C*-Algebra Equal to Norm

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $r_A$ be the spectral radius in $A$.
Let $x \in A$ be normal.
Then $\map {r_A} x = \norm x$. 
=== Corollary ===
== Proof ==
We first take $x$ Hermitian.
From the $\text C^\ast$ identity, we have:
:$\norm {x^2} = \norm {x x^\ast} = \norm x^2$
for each $x \in A$.
=== Lemma ===
Hence, we have:
:$\norm {x^{2^n} }^{2^{-n} } = \norm x$ 
Hence:
:$\ds \lim_{n \mathop \to \infty} \norm {x^{2^n} }^{2^{-n} } = \norm x$ 
From Gelfand's Spectral Radius Formula: Banach Algebra, we have:
:$\ds \map {r_A} x = \lim_{n \mathop \to \infty} \norm {x^n}^{1/n}$
Hence from Limit of Subsequence equals Limit of Sequence, we have:
:$\ds \map {r_A} x = \lim_{n \mathop \to \infty} \norm {x^{2^n} }^{2^{-n} }$
Hence we obtain:
:$\map {r_A} x = \norm x$ 
when $x$ is Hermitian.
Now take $x$ general normal.
From Product of Element in *-Star Algebra with its Star is Hermitian, $x x^\ast$ is Hermitian.
Hence we have:
:$\map {r_A} {x x^\ast} = \norm {x x^\ast}$
By the $\text C^\ast$ identity we then have:
:$\map {r_A} {x x^\ast} = \norm x^2$
From Spectral Radius is Multiplicative on Commuting Elements in Banach Algebra, we have:
:$\map {r_A} {x x^\ast} \le \map {r_A} x \map {r_A} {x^\ast}$ 
From Spectrum of Element of Banach Algebra is Bounded: Corollary, we have:
:$\map {r_A} x \le \norm x$
Hence if $x = {\mathbf 0}_A$, we have $\map {r_A} x = 0$ and hence the theorem. 
Now take $x \ne {\mathbf 0}_A$. 
We have:
:$\norm x^2 \le \map {r_A} x \map {r_A} {x^\ast} \le \norm {x^\ast} \map {r_A} x = \norm x \map {r_A} x$
Dividing by $\norm x$, we have:
:$\norm x \le \map {r_A} x$
Since also:
:$\map {r_A} x \le \norm x$
we have:
:$\map {r_A} x = \norm x$
Category:C*-Algebras
Category:Normal Elements of *-Algebras
Category:Spectral Radius
Category:Spectral Radius of Normal Element of C*-Algebra Equal to Norm
Spectral Radius of Normal Element of C*-Algebra Equal to Norm/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $r_A$ be the spectral radius in $A$.
Let $x \in A$ be normal such that:
:$\map {r_A} x = 0$
Then $x = {\mathbf 0}_A$.
== Proof ==
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have:
:$\norm x = \map {r_A} x = 0$
Hence from , we have $x = {\mathbf 0}_A$.
Category:Spectral Radius of Normal Element of C*-Algebra Equal to Norm
Spectral Radius of Normal Element of C*-Algebra Equal to Norm/Lemma

== Lemma ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $x \in A$ be Hermitian.
Then:
:$\norm {x^{2^n} } = \norm x^{2^n}$ 
for each $n \in \N$. 
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\norm {x^{2^n} } = \norm x^{2^n}$ for all Hermitian $x \in A$. 
=== Basis for the Induction ===
We have:
:$\norm {x x^\ast} = \norm x^2$ for all Hermitian $x \in A$
from the definition of a $\text C^\ast$-algebra. 
Since $x$ is Hermitian, we have:
:$x^\ast = x$
and hence:
:$\norm {x^2} = \norm x^2$ for all Hermitian $x \in A$.
Hence $\map P 1$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\norm {x^{2^k} } = \norm x^{2^k}$ for all Hermitian $x \in A$.
Then we need to show:
:$\norm {x^{2^{k + 1} } } = \norm x^{2^{k + 1} }$ for all Hermitian $x \in A$.
=== Induction Step ===
This is our induction step.
For each Hermitian $x \in A$, we have:
while from the base case, we have:
:$\norm {x^2}^{2^k} = \paren {\norm x^2}^{2^k} = \norm x^{2 \cdot 2^k} = \norm x^{2^{k + 1} }$
Hence we have:
:$\norm {x^{2^{k + 1} } } = \norm x^{2^{k + 1} }$ for all Hermitian $x \in A$.
Category:Spectral Radius of Normal Element of C*-Algebra Equal to Norm
Spectral Theorem for Compact Hermitian Operators

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $T \in \map {B_0} \HH$ be a compact Hermitian operator.
Then there exists a (possibly finite) sequence $\sequence {\lambda_n}$ of distinct nonzero eigenvalues of $T$ such that:
:$(1): \quad P_n P_m = P_m P_n = 0$ if $n \ne m$
:$(2): \quad \ds \lim_{k \mathop \to \infty} \norm {T - \sum_{n \mathop = 1}^k \lambda_n P_n}$, that is, $T = \ds \sum_{n \mathop = 1}^\infty \lambda_n P_n$
where:
:$P_n$ is the orthogonal projection onto the eigenspace of $\lambda_n$
:$\norm {\, \cdot \,}$ denotes the norm on bounded linear operators.
=== Corollary 1 ===
There exists a (possibly finite) sequence $\sequence {\mu_n}$ of real numbers and a basis $E = \sequence {e_n}$ for $\paren {\ker T}^\perp$ such that:
:$\forall h \in H: T h = \ds \sum_{n \mathop = 1}^\infty \innerprod h {e_n}_\HH \mu_n e_n$
=== Corollary 2 ===
If $T$ has trivial kernel, then $\HH$ is finite dimensional.
== Proof ==
== Also see ==
* Spectral Theorem for Compact Normal Operators, a more general result
== Sources ==
* : $\text {II}.5.1-5, 9$
Category:Hermitian Operators
Category:Spectral Theory
Category:Spectra (Bounded Linear Operators)
Spectrum of Adjoint of Bounded Linear Operator

== Theorem ==
Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator.
Let $T^\ast : X \to X$ be the adjoint of $T$. 
Let $\map \sigma T$ and $\map \sigma {T^\ast}$ be the spectrum of $T$ and $T^\ast$ respectively. 
Then: 
:$\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$ 
where $\overline \lambda$ denotes the complex conjugate of $\lambda$. 
== Proof ==
We show that for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ $\overline \lambda \not \in \map \sigma {T^\ast}$.
Let $\lambda \in \C$ have $\lambda \not \in \map \sigma T$.
Then $T - \lambda I$ is invertible as a bounded linear operator.
Note that $I^\ast = I$ from Adjoint of Identity Transformation. 
From Adjoining is Linear, we have that $T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$.
From Adjoining Commutes with Inverting, we have: 
:$T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$ is invertible
So we have $\overline \lambda \not \in \map \sigma {T^\ast}$. 
So if $\lambda \not \in \map \sigma T$ then $\overline \lambda \not \in \map \sigma {T^\ast}$. 
From Adjoint is Involutive, swapping $T$ for $T^\ast$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$. 
From Complex Conjugation is Involution, swapping $\lambda$ for $\overline \lambda$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$. 
So for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ $\overline \lambda \not \in \map \sigma {T^\ast}$.
So we have: 
:$\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$ 
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Spectra (Bounded Linear Operators)
Category:Adjoints
Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space/Corollary

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $\struct {\Phi_A, w^\ast}$ be the spectrum of $A$. 
Then $\struct {\Phi_A, w^\ast}$ is a compact Hausdorff space. 
== Proof ==
Let $\phi \in \Phi_A$. 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have that:
:the condition that $\map \phi x \map \phi y = \map \phi {x y}$ for all $x, y \in A$ implies that $\map \phi { {\mathbf 1}_A} = 1$.
That is, $\map \phi x \map \phi y = \map \phi {x y}$ for all $x, y \in A$ implies that $\phi \ne {\mathbf 0}_{A^\ast}$. 
So, we have:
:$\Phi_A = \set {\phi \in A^\ast : \map \phi x \map \phi y = \map \phi {x y} \text { for all } x, y \in A}$
In the proof of Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space, it was established that $\struct {\set {\phi \in A^\ast : \map \phi x \map \phi y = \map \phi {x y} \text { for all } x, y \in A}, w^\ast}$ is compact and Hausdorff.
Hence $\struct {\Phi_A, w^\ast}$ is a compact Hausdorff space.
Category:Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space
Spectrum of Bounded Linear Operator contains Point Spectrum

== Theorem ==
Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator. 
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Let $\map \sigma T$ be the spectrum of $T$. 
Then $\map {\sigma_p} T \subseteq \map \sigma T$.
== Proof ==
Let $\lambda \in \map {\sigma_p} T$. 
Then there exists $x \ne \mathbf 0_X$ such that $T x = \lambda x$.
So $\paren {T - \lambda I} x = 0$ for some $x \ne \mathbf 0_X$. 
So $\ker T \ne \set {\mathbf 0_X}$.
So from Linear Transformation is Injective iff Kernel Contains Only Zero, $T$ is not injective. 
So $T$ cannot be invertible as a bounded linear operator.
So $\lambda \in \map \sigma T$.
Category:Point Spectra (Linear Operators)
Category:Spectra (Bounded Linear Operators)
Spectrum of Bounded Linear Operator equal to Spectrum as Densely-Defined Linear Operator

== Theorem ==
Let $\HH$ be a Hilbert space over $\C$. 
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map {\sigma_1} T$ be the spectrum of $T$ as a bounded linear operator.
Let $\map {\sigma_2} T$ be the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
Then: 
:$\map {\sigma_1} T = \map {\sigma_2} T$
 
== Proof ==
From the definition of the spectrum of $T$ as a bounded linear operator, we have: 
:$\map {\sigma_1} T = \C \setminus \map {\rho_1} T$
where $\map {\rho_1} T$ is the resolvent set of $T$ as a bounded linear operator.
From the definition of the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$, we have: 
:$\map {\sigma_2} T = \C \setminus \map {\rho_2} T$
where $\map {\rho_2} T$ is the resolvent set of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
From Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator, we have: 
:$\map {\rho_1} T = \map {\rho_2} T$
So:
:$\map {\sigma_1} T = \map {\sigma_2} T$
Category:Spectra (Bounded Linear Operators)
Category:Spectra (Densely-Defined Linear Operators)
Spectrum of Bounded Linear Operator is Closed

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\C$..
Let $T$ be a bounded linear operator on $X$.
Then the spectrum $\map \sigma T$ of $T$ is a closed set in $\C$.
== Proof ==
From Resolvent Set of Bounded Linear Operator is Open, the resolvent set $\map \rho T$ is open. 
From the definition of spectrum, we have $\map \sigma T = \C \setminus \map \rho T$.
From the definition of a closed set, $\map \sigma T$ is closed set in $\C$.
Category:Bounded Linear Operators
Category:Spectra (Bounded Linear Operators)
Spectrum of Bounded Linear Operator is Compact

== Theorem ==
Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator. 
Let $\map \sigma T$ be the spectrum of $T$. 
Then $\map \sigma T$ is compact, and: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
 
== Proof ==
From Spectrum of Bounded Linear Operator is Closed, $\map \sigma T$ is closed in $\C$.
It therefore suffices to show that $\map \sigma T$ is bounded.
Hence it will suffice to show that: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
Let $\lambda \in \C$ be such that $\cmod \lambda > \norm T_{\map \BB X}$. 
Then, we have: 
:$\ds \norm {\frac 1 \lambda T}_{\map \BB X} = \frac 1 {\cmod \lambda} \norm T_{\map \BB X} < 1$
from .
From Invertibility of Identity Minus Operator:
:$\ds I - \frac 1 \lambda T$ is invertible as a bounded linear operator.
Since $\map \BB X$ is a vector space, we have: 
:$\ds -\frac 1 \lambda \paren {I - \frac 1 \lambda T} = T - \lambda I$ is invertible as a bounded linear operator.
That is $\lambda \in \map \rho T$, where $\map \rho T$ is the resolvent set of $T$. 
So if $\cmod \lambda > \norm T_{\map \BB X}$, we have $\lambda \in \C \setminus \map \sigma T$. 
So if $\lambda \in \map \sigma T$, we have $\cmod \lambda \le \norm T_{\map \BB X}$.
That is: 
:$\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Spectra (Bounded Linear Operators)
Category:Compact Spaces
Spectrum of Bounded Linear Operator is Non-Empty

== Theorem ==
Let $B$ be a Banach space over $\C$.
Let $\map {\mathfrak L} {B, B}$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \map {\mathfrak L} {B, B}$.
Then the spectrum $\map \sigma T$ of $T$ is non-empty.
== Proof ==
Let $f: \C \to \map {\mathfrak L} {B, B}$ be the resolvent mapping defined as $\map f z = \paren {T - z I}^{-1}$.
 the spectrum of $T$ is empty, so that $\map f z$ is well-defined for all $z \in \C$. 
We first show that $\norm {\map f z}_*$ is uniformly bounded by some constant $C$.
Observe that:
:$(1): \quad \norm {\map f z}_* = \norm {\paren {T - z I}^{-1} }_* = \dfrac 1 {\size z} \norm {\paren {I - \dfrac T z}^{-1} }_*$
For $\size z \ge 2 \norm T_*$, Operator Norm is Norm implies that:
:$\norm {\dfrac T z}_* \le \dfrac {\norm T_*} {2 \norm T_*} = \dfrac 1 2$
Hence by $(1)$ and Invertibility of Identity Minus Operator, we get:
Therefore, the norm of $\map f z$ is bounded for $\size z \ge 2 \norm T_*$ by some constant $C_1$.
Next, consider the disk $\size z \le 2 \norm T_*$ in the complex plane.
It is compact.
From Resolvent Mapping is Continuous $\map f z$ is continuous on the disk.
From Norm is Continuous, we get from Continuous Function on Compact Space is Bounded that $\norm f_*$ is bounded on the this disk by some constant $C_2$.
Thus, $\norm {\map f z}_*$ is bounded for all $z \in \C$ by $C = \max \set {C_1, C_2}$.
Finally, pick any $x \in B$ and $\ell \in B^*$, the dual of $B$.
Define the function $g: \C \to \C$ as:
:$\forall z \in \C: \map g z = \map \ell {\map f z x}$
Since $f$ has empty spectrum, Resolvent Mapping is Analytic and Strongly Analytic iff Weakly Analytic together imply that $g$ is an entire function.
Thus we have:
So $g$ is a bounded entire function.
It is therefore equal to some constant $K$ by Liouville's Theorem.
But the inequality above:
:$\size {\map g z} \le \norm \ell_{B^*} \norm {\paren {T - z I}^{-1} }_* \norm x_B$
together with Resolvent Mapping Converges to 0 at Infinity, implies that:
:$\ds \size K = \lim_{z \mathop \to \infty} \size {\map g z} \le 0$
So $g$ is the constant function $0$.
We have therefore shown that:
:$\forall x \in B, \ell \in B^*: \map \ell {\map f z x} = 0$
This implies from Condition for Bounded Linear Operator to be Zero that $\map f z = 0$.
In particular:
:$\map f 0 = T^{-1} = 0$
But this is a contradiction, since our assumption that the spectrum of $T$ is empty implies that $T$ has a two-sided bounded inverse.
Category:Bounded Linear Operators
Category:Spectra (Bounded Linear Operators)
Spectrum of Bounded Linear Operator on Finite-Dimensional Banach Space is equal to Point Spectrum

== Theorem ==
Let $X$ be a finite-dimensional Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\map \sigma T = \map {\sigma_p} T$. 
== Proof ==
We have that $\lambda \in \map \sigma T$ $T - \lambda I$ is not invertible as a bounded linear transformation.
So $T - \lambda I$ is not bijective or its inverse $\paren {T - \lambda I}^{-1}$ is not bounded.
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, every linear operator $S : X \to X$ is bounded.
So $T - \lambda I$ is not invertible as a bounded linear transformation $T - \lambda I$ is not bijective.
From Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective, we have that $T - \lambda I$ is not bijective $T - \lambda I$ is not injective. 
From Linear Transformation is Injective iff Kernel Contains Only Zero, we have $T - \lambda I$ is not injective $\ker T \ne \set {\mathbf 0_X}$.
That is, there exists $x \ne \mathbf 0_X$ such that: 
:$\paren {T - \lambda I} x = 0$
and hence $T x = \lambda x$. 
So $\lambda \in \map {\sigma_p} T$. 
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Spectra (Bounded Linear Operators)
Category:Point Spectra (Linear Operators)
Category:Finite Dimensional Vector Spaces
Spectrum of Commutative C*-Algebra is Non-Empty

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a commutative $\text C^\ast$-algebra such that:
:$A \ne \set { {\mathbf 0}_A}$
where ${\mathbf 0}_A$ is the zero vector of $A$.
Let $\Phi_A$ be the spectrum of $A$.
Then $\Phi_A \ne \O$. 
== Proof ==
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $r_A$ denotes spectral radius in $A$.
From Non-Zero C*-Algebra contains Non-Zero Hermitian Element:
:there exists a Hermitian element $x \in A \setminus \set { {\mathbf 0}_A}$.
Hence from , we have $\norm a \ne 0$. 
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have:
:$\norm a = \map {r_A} a = \sup \set {\cmod \lambda : \lambda \in \map {\sigma_A} a} \ne 0$
Hence there exists $\lambda \in \map {\sigma_A} a \setminus \set 0$. 
From Spectrum of Element of Unital Commutative Banach Algebra and , we have:
:$\map {\sigma_A} a \subseteq \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$
Then:
:$\lambda \in \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$
while $\lambda \ne 0$.
So there exists $\phi \in \Phi_A$ such that $\map \phi x = \lambda$.
In particular $\Phi_A \ne \O$.
Category:C*-Algebras