nyuuzyou/wikis huggingface dataset:
Spectrum of Element of Unital Commutative Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital 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 $x$. 
Then:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$ 
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
We have $\lambda \in \map {\sigma_A} x$ :
:$\lambda {\mathbf 1}_A - x$ is not invertible.
From Element of Unital Commutative Banach Algebra is Invertible iff not in Kernel of Character, this is the case there exists a character $\phi$ such that:
:$\map \phi {\lambda {\mathbf 1}_A - x} = 0$
From linearity, we have:
:$\map \phi x = \lambda \map \phi { {\mathbf 1}_A}$ 
in this case. 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$.
Hence we have:
:$\lambda \in \map {\sigma_A} x$ there exists a character $\phi$ with $\lambda = \map \phi x$.
This is precisely the statement:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$ 
Category:Commutative Banach Algebras
Category:Unital Banach Algebras
Category:Spectrum of Element of Unital Commutative Banach Algebra
Spectrum of Element of Unital Commutative Banach Algebra/Corollary 1

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$ that is not unital as an algebra. 
Let $\Phi_A$ be the spectrum of $A$.
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$. 
Then:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$ 
== Proof ==
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $\struct {A, \norm {\, \cdot \,} }$.
By the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
Let $\Phi_{A_+}$ be the spectrum of $A_+$.
From Unitization of Commutative Algebra over Field is Commutative, $A_+$ is commutative.
Hence by Spectrum of Element of Unital Commutative Banach Algebra we have:
:$\map {\sigma_A} x = \set {\map \phi {x, 0} : \phi \in \Phi_{A_+} }$
From Characterization of Character on Banach Algebra, we have:
:$\Phi_{A_+} = \set {\widetilde \phi : \phi \in \Phi_A} \cup \set \psi$
where:
:$\map {\widetilde \phi} {x, \lambda} = \map \phi x + \lambda$
and:
:$\map \psi {x, \lambda} = \lambda$
for each $\tuple {x, \lambda} \in A_+$. 
Hence we have:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set {\map \psi 0} = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$.
Category:Spectrum of Element of Unital Commutative Banach Algebra
Spectrum of Element of Unital Commutative Banach Algebra/Corollary 2

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $\Phi_A$ be the spectrum of $A$.
Then $\Phi_A \ne \O$.
== Proof ==
From Spectrum of Element of Banach Algebra is Non-Empty, we have $\map {\sigma_A} x \ne \O$.
That is, there exists $\lambda \in \map {\sigma_A} x$.
From Spectrum of Element of Unital Commutative Banach Algebra, there exists a character $\phi$ with $\map \phi x = \lambda$.
In particular, $\phi \in \Phi_A$.
So $\Phi_A \ne \O$.
Category:Spectrum of Element of Unital Commutative Banach Algebra
Spectrum of Hermitian Element in Unital C*-Algebra is Real

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be Hermitian. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x \subseteq \R$. 
== Proof ==
Let $A' \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $A'$ is commutative.
Let $\Phi_{A'}$ be the spectrum of $A'$.
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_{A'} } x$
From Spectrum of Element of Unital Commutative Banach Algebra, we have $\map {\sigma_A} x = \map {\sigma_{A'} } x = \set {\map \phi x : \phi \in \Phi_{A'} }$. 
From Character on Unital C*-Algebra is Real at Hermitian Elements, we have $\map \phi x \in \R$ for each $\phi \in \Phi_{A'}$. 
Hence $\map {\sigma_A} x \subseteq \R$. 
Category:C*-Algebras
Category:Hermitian Elements of *-Algebras
Spectrum of Idempotent in Algebra over Complex Numbers

== Theorem ==
Let $A$ be an algebra over $\C$.
Let $p \in A$ be idempotent, that is:
:$p^2 = p$
Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$. 
Then:
:$\map {\sigma_A} p \subseteq \set {0, 1}$
== Proof ==
Suppose first that $A$ is not unital and let $p \in A$ be idempotent.
Let $A_+$ be the unitization of $A$. 
Then we have:
So $\tuple {p, 0}$ is an idempotent in $A_+$.
Since, by definition of the spectrum in a non-unital algebra, we have:
:$\map {\sigma_A} p = \map {\sigma_{A_+} } {\tuple {p, 0} }$
with $A_+$ unital and $\tuple {p, 0}$ idempotent.
 we can therefore take $A$ to be unital.
Let $\lambda \in \C$ be such that $\lambda \not \in \set {0, 1}$.
That is, such that $\lambda \paren {1 - \lambda} \ne 0$.
We have:
so that:
:$\ds \paren {\lambda {\mathbf 1}_A - p} \paren {\frac {\paren {1 - \lambda} {\mathbf 1}_A - p} {\lambda \paren {1 - \lambda} } } = {\mathbf 1}_A$
Since $p$ and ${\mathbf 1}_A$ commute, we also have:
:$\ds \paren {\frac {\paren {1 - \lambda} {\mathbf 1}_A - p} {\lambda \paren {1 - \lambda} } } \paren {\lambda {\mathbf 1}_A - p} = {\mathbf 1}_A$
Hence:
:$\lambda {\mathbf 1}_A - p \in \map G A$ for $\lambda \not \in \set {0, 1}$
where $\map G A$ is the group of units of $A$.
Hence, we have:
:$\map {\sigma_A} p = \set {\lambda \in \C : \lambda {\mathbf 1}_A - p \not \in \map G A} \subseteq \set {0, 1}$
== Sources ==
* : $4.5$: Proposition $4.16$
Category:Spectra (Spectral Theory)
Spectrum of Identity Element of Unital Algebra

== Theorem ==
Let $A$ be a unital algebra over $\C$ with at least two distinct elements.
Let $\sigma_A$ be the spectrum in $A$.
Then we have:
:$\map {\sigma_A} { {\mathbf 1}_A} = \set 1$
== Proof ==
We have:
:${\mathbf 1}_A - {\mathbf 1}_A = {\mathbf 0}_A$
which is not invertible.
So $1 \in \map {\sigma_A} { {\mathbf 1}_A}$.
Now suppose that $\lambda \in \C \setminus \set 1$. 
Then we have:
:$\lambda {\mathbf 1}_A - {\mathbf 1}_A = \paren {\lambda - 1} {\mathbf 1}_A$.
Hence:
:$\paren {\paren {\lambda - 1} {\mathbf 1}_A} \paren {\paren {\lambda - 1}^{-1} {\mathbf 1}_A} = {\mathbf 1}_A$
and:
:$\paren {\paren {\lambda - 1}^{-1} {\mathbf 1}_A} \paren {\paren {\lambda - 1} {\mathbf 1}_A} = {\mathbf 1}_A$
So $\paren {\lambda - 1} {\mathbf 1}_A$ is invertible with inverse $\paren {\lambda - 1}^{-1} {\mathbf 1}_A$.
Hence we have $\lambda \not \in \map {\sigma_A} { {\mathbf 1}_A}$ for $\lambda \in \C \setminus 1$. 
So we have $\map {\sigma_A} { {\mathbf 1}_A} = \set 1$.
Category:Spectra (Spectral Theory)
Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism

== Theorem ==
Let $A, B$ be unital algebras over $\C$.
Let $\phi : A \to B$ be a unital algebra homomorphism.
Let $x \in A$.
We then have:
:$\map {\sigma_B} {\map \phi x} \subseteq \map {\sigma_A} x$ 
where $\sigma_A$ and $\sigma_B$ denote spectra in $A$ and $B$ respectively. 
If $\phi$ is a unital algebra isomorphism, we further have $\map {\sigma_B} {\map \phi x} = \map {\sigma_A} x$. 
=== Corollary ===
== Proof ==
We show that:
:$\map {\rho_A} x \subseteq \map {\rho_B} {\map \phi x}$
where $\rho_A$ and $\rho_B$ denote the resolvent sets in $A$ and $B$ respectively.
Let $\lambda \in \map {\rho_A} x$.
Then there exists $y \in A$ such that $\paren {\lambda {\mathbf 1}_A - x} y = y \paren {\lambda {\mathbf 1}_A - x} = {\mathbf 1}_A$.
Hence we have:
:$\map \phi {\paren {\lambda {\mathbf 1}_A - x} y} = \map \phi {y \paren {\lambda {\mathbf 1}_A - x} } = \map \phi { {\mathbf 1}_A} = {\mathbf 1}_B$
since $\phi$ is a unital algebra homomorphism.
Since $\phi$ is an algebra homomorphism, we have:
:$\map \phi {\lambda {\mathbf 1}_A - x} \map \phi y = \map \phi y \map \phi {\paren {\lambda {\mathbf 1}_A - x} } = {\mathbf 1}_B$
Hence $\map \phi y$ is an inverse for $\map \phi {\lambda {\mathbf 1}_A - x}$.
Since $\phi$ is an algebra homomorphism, we have $\map \phi {\lambda {\mathbf 1}_A - x} = \lambda {\mathbf 1}_B - \map \phi x$. 
Hence $\lambda \in \map {\rho_B} {\map \phi x}$. 
So we have:
:$\map {\rho_A} x \subseteq \map {\rho_B} {\map \phi x}$
Hence:
:$\map {\sigma_B} {\map \phi x} \subseteq \map {\sigma_A} x$ 
If $\phi$ is a unital algebra isomorphism, we have that $\phi^{-1} : B \to A$ is also a unital algebra isomorphism from Inverse of Algebraic Structure Isomorphism is Isomorphism: General Result.
Hence we have:
:$\map {\sigma_A} {\map {\phi^{-1} } y} \subseteq \map {\sigma_B} y$ 
for each $y \in B$. 
For $x \in A$, we can set $y = \map \phi x$ to obtain:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} {\map \phi x}$
We therefore obtain:
:$\map {\sigma_A} x = \map {\sigma_B} {\map \phi x}$ 
Category:Spectra (Spectral Theory)
Category:Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism
Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism/Corollary

== Theorem ==
Let $A, B$ be algebras over $\C$.
Let $\phi : A \to B$ be an algebra homomorphism.
Let $x \in A$.
Then:
:$\map {\sigma_B} {\map \phi x} \cup \set 0 \subseteq \map {\sigma_A} x \cup \set 0$
where $\sigma_A$ and $\sigma_B$ denote spectra in $A$ and $B$ respectively. 
== Proof ==
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
Define $\psi : A_+ \to B_+$ by:
:$\map \psi {a, \lambda} = \tuple {\map \phi a, \lambda}$ 
for each $\tuple {a, \lambda} \in A_+$.
From Induced Algebra Homomorphism on Unitization, $\psi$ is a unital algebra homomorphism.
Hence we have:
:$\map {\sigma_{B_+} } {\tuple {\map \phi x, 0} } = \map {\sigma_{B_+} } {\map \psi {x, 0} } \subseteq \map {\sigma_{A_+} } {\tuple {x, 0} }$
From Spectrum of Element in Unitization of Unital Algebra, we have:
:$\map {\sigma_{B_+} } {\tuple {\map \phi x, 0} } = \map {\sigma_B} {\map \phi x} \cup \set 0$
and:
:$\map {\sigma_{A_+} } {\tuple {x, 0} } = \map {\sigma_A} x \cup \set 0$
Hence:
:$\map {\sigma_B} {\map \phi x} \cup \set 0 \subseteq \map {\sigma_A} x \cup \set 0$
Category:Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism
Spectrum of Inverse of Element of Unital Algebra

== Theorem ==
Let $A$ be an algebra over $\C$.
Let $a \in A$ be invertible.
Let $\sigma_A$ denote the spectrum in $A$.
Then we have:
:$\map {\sigma_A} {a^{-1} } = \set {\lambda^{-1} : \lambda \in \map {\sigma_A} a}$
== Proof ==
Since $a$ is invertible, we have $0 \not \in \map {\sigma_A} a$ and $0 \not \in \map {\sigma_A} {a^{-1} }$.
We show that for $\lambda \in \C \setminus \set 0$:
:$\lambda {\mathbf 1}_A - a$ is invertible $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible.
We can write:
:$\lambda^{-1} {\mathbf 1}_A - a^{-1} = a^{-1} \paren {\lambda^{-1} a - {\mathbf 1}_A} = -\lambda^{-1} a^{-1} \paren {\lambda {\mathbf 1}_A - a}$
Hence if $\lambda {\mathbf 1}_A - a$ is invertible, we have that $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible from Inverse of Product.
Similarly we have:
:$\lambda {\mathbf 1}_A - a = -\lambda a \paren {\lambda^{-1} {\mathbf 1}_A - a^{-1} }$
Hence if $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible then from Inverse of Product we also obtain that $\lambda {\mathbf 1}_A - a$ is invertible.
So:
:$\lambda {\mathbf 1}_A - a$ is invertible $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible for $\lambda \ne 0$.
We have that $0$ is in neither $\map {\sigma_A} a$ nor $\map {\sigma_A} {a^{-1} }$ as previously pointed out.
Hence we have $\lambda \in \map {\sigma_A} a$ $\lambda^{-1} \in \map {\sigma_A} {a^{-1} }$ for all $\lambda \in \C$.
This is precisely the statement that $\map {\sigma_A} {a^{-1} } = \set {z^{-1} : z \in \map {\sigma_A} a}$.
Category:Spectra (Spectral Theory)
Spectrum of Product of Elements of Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $\sigma_A$ denote spectrum in $A$.
Let $x, y \in A$.
Then we have:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$ 
== Proof ==
First take $A$ to be unital.
We show that ${\mathbf 1}_A - x y$ is invertible ${\mathbf 1}_A - y x$ is invertible. 
Swapping $x$ and $y$ it suffices to show that if ${\mathbf 1}_A - x y$ is invertible then ${\mathbf 1}_A - y x$ is invertible.
Suppose that ${\mathbf 1}_A - x y$ is invertible.
We have:
and:
So ${\mathbf 1}_A - y x$ is invertible with inverse ${\mathbf 1}_A y \paren { {\mathbf 1}_A - x y}^{-1} x$.
Hence ${\mathbf 1}_A - x y$ is invertible ${\mathbf 1}_A - y x$ is invertible.
Now take $\lambda \in \C \setminus \set 0$. 
Take $\mu \in \C \setminus \set 0$ such that $\lambda = \mu^2$. 
Then applying the above, we have that ${\mathbf 1}_A - \paren {x/\mu} \paren {y/\mu}$ is invertible ${\mathbf 1}_A - \paren {y/\mu} \paren {x/\mu}$ is invertible.
Hence ${\mathbf 1}_A - \lambda^{-1} x y$ is invertible ${\mathbf 1}_A - \lambda^{-1} y x$ is invertible if $\lambda \ne 0$. 
Hence $\lambda {\mathbf 1}_A - x y$ is invertible $\lambda {\mathbf 1}_A - y x$ invertible if $\lambda \ne 0$. 
This is precisely the statement:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x}$
Now suppose that $A$ is not unital.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By the unital case we have:
:$\map {\sigma_{A_+} } {\tuple {x, 0} \tuple {y, 0} } \setminus \set 0 = \map {\sigma_{A_+} } {\tuple {y, 0} \tuple {x, 0} } \setminus \set 0$
That is:
:$\map {\sigma_{A_+} } {\tuple {x y, 0} } \setminus \set 0 = \map {\sigma_{A_+} } {\tuple {y x, 0} } \setminus \set 0$
Hence by the definition of the spectrum in a non-unital algebra:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$ 
Category:Spectrum (Spectral Theory)
Category:Banach Algebras
Spectrum of Projection in *-Algebra

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $p$ be a projection on $A$.
Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$.
Then $\map {\sigma_A} p \subseteq \set {0, 1}$.
=== Corollary ===
== Proof ==
From the definition of a projection, we have $p^2 = p$.
Hence $p^2 - p = {\mathbf 0}_A$.
Hence from Spectrum of Zero Vector in Algebra, we have $\map {\sigma_A} {p^2 - p} = \map {\sigma_A} { {\mathbf 0}_A} = \set 0$.
From the Spectral Mapping Theorem for Polynomials, we have:
:$\set 0 = \set {z^2 - z : z \in \map {\sigma_A} p}$
In other words:
:if $z \in \map {\sigma_A} p$, then $z^2 - z = 0$.
That is:
:$z = 0$ or $z = 1$.
So we obtain $\map {\sigma_A} p \subseteq \set {0, 1}$.
Category:Projections (*-Algebras)
Category:Spectra (Spectral Theory)
Category:Spectrum of Projection in *-Algebra
Spectrum of Projection in *-Algebra/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $p$ be a projection on $A$.
Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$.
Then:
:$p = {\mathbf 0}_A$ $\map {\sigma_A} p = \set 0$
:$p = {\mathbf 1}_A$ $\map {\sigma_A} p = \set 1$
:$p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$ $\map {\sigma_A} p = \set {0, 1}$. 
== Proof ==
Note first that since $p$ is Hermitian, it is normal.
=== $p = {\mathbf 0}_A$ $\map {\sigma_A} p = \set 0$ ===
From Spectrum of Zero Vector in Algebra, we have $\map {\sigma_A} p = \set 0$ if $p = {\mathbf 0}_A$.
Conversely, if $\map {\sigma_A} p = \set 0$ then $p = {\mathbf 0}_A$ from Spectral Radius of Normal Element of C*-Algebra Equal to Norm: Corollary.
=== $p = {\mathbf 1}_A$ $\map {\sigma_A} p = \set 1$ ===
From Spectrum of Identity Element of Unital Algebra, we have $\map {\sigma_A} p = \set 1$ if $p = {\mathbf 0}_A$.
Now suppose that $\map {\sigma_A} p = \set 1$. 
From Spectral Mapping Theorem for Polynomials we have $\map {\sigma_A} {p - {\mathbf 1}_A} = \set 0$.
Then from Spectral Radius of Normal Element of C*-Algebra Equal to Norm: Corollary, we have $p - {\mathbf 1}_A = {\mathbf 0}_A$ and hence $p = {\mathbf 1}_A$. 
=== $p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$ $\map {\sigma_A} p = \set {0, 1}$ ===
This is essentially putting the last two cases together.
From Spectrum of Projection in *-Algebra, we have $\map {\sigma_A} p \subseteq \set {0, 1}$. 
From Spectrum of Element of Banach Algebra is Non-Empty, we have $\map {\sigma_A} p \ne \O$.
So $\map {\sigma_A} p = \set 0$, $\map {\sigma_A} p = \set 1$ or $\map {\sigma_A} p = \set {0, 1}$. 
If $p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$, then $\map {\sigma_A} p \ne \set 0$ by the first case and $\map {\sigma_A} p \ne \set 1$ in the second.
So we have $\map {\sigma_A} p = \set {0, 1}$ if $p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$.
Conversely, if $\map {\sigma_A} p = \set {0, 1}$ then since $\map {\sigma_A} p \ne \set 0$ we have $p \ne {\mathbf 0}_A$.
Similarly since $\map {\sigma_A} p \ne \set 1$, we have $p \ne {\mathbf 1}_A$. 
Hence $p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$.
Category:Spectrum of Projection in *-Algebra
Spectrum of Ring is Nonempty

== Theorem ==
Let $A$ be a non-trivial commutative ring with unity.
Then its prime spectrum is non-empty:
:$\Spec A \ne \O$
== Proof ==
This is a reformulation of Ring with Unity has Prime Ideal.
== Also see ==
* Maximal Spectrum of Ring is Nonempty
Category:Commutative Algebra
Spectrum of Self-Adjoint Bounded Linear Operator is Real

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $T : \HH \to \HH$ be a bounded self-adjoint operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then:
:$\map \sigma T \subseteq \R$
== Proof 1 ==
== Proof 2 ==
Category:Hermitian Operators
Category:Spectra (Bounded Linear Operators)
Spectrum of Self-Adjoint Bounded Linear Operator is Real/Proof 1

== Theorem ==
== Proof ==
This follows from:
:Spectrum of Self-Adjoint Densely-Defined Linear Operator is Real and Closed
:Spectrum of Bounded Linear Operator equal to Spectrum as Densely-Defined Linear Operator
Category:Hermitian Operators
Category:Spectra (Bounded Linear Operators)
Spectrum of Self-Adjoint Bounded Linear Operator is Real/Proof 2

== Theorem ==
== Proof ==
Let $\lambda := a + i b \in \C \setminus \R$.
Note that $b \ne 0$.
For all $\phi \in \HH$:
In view of $(1)$, both $T - \lambda I$ and $T - \overline \lambda I$ are injective.
Moreover:
By Linear Subspace Dense iff Zero Orthocomplement, $\Img {T - \lambda I}$ is dense in $\HH$.
Let $\phi \in \HH$.
Then there exists an $\sequence {\psi _n} \subseteq \HH$ such that:
:$\paren {T - \lambda I} \psi_n \to \phi$
By $(1)$ there exists a $\psi \in \HH$ such that:
:$\psi_n \to \psi$
and:
:$\paren {T - \lambda I} \psi = \phi$
Thus $T - \lambda I$ is surjective.
In addition, by $(1)$ we have:
:$\norm \psi \le \size b^{-1} \norm \phi$
Therefore $T - \lambda I$ is invertible such that:
:$\norm {\paren {T - \lambda I}^{-1} } \le \size b^{-1}$
That is, $\lambda \not \in \map \sigma T$.
Category:Hermitian Operators
Category:Spectra (Bounded Linear Operators)
Spectrum of Star of Element in *-Algebra

== Theorem ==
Let $\struct {A, \ast}$ be a unital $\ast$-algebra over $\C$.
Let $a \in A$. 
Let $\sigma_A$ denote the spectrum.
Then:
:$\map {\sigma_A} {a^\ast} = \set {\overline \lambda : \lambda \in \map {\sigma_A} a}$
== Proof ==
From the definition of an involution, we have:
:$\paren {\lambda {\mathbf 1}_A - a}^\ast = \overline \lambda {\mathbf 1}_A^\ast - a^\ast$ 
From Identity Element in Unital *-Algebra is Hermitian, we therefore have:
:$\paren {\lambda {\mathbf 1}_A - a}^\ast = \overline \lambda {\mathbf 1}_A - a^\ast$ 
From Inverse of Star of Element in Unital *-Algebra: Corollary, we deduce that:
:$\lambda {\mathbf 1}_A - a$ is invertible $\overline \lambda {\mathbf 1}_A - a^\ast$ is invertible.
That is to say:
:$\lambda \in \map {\sigma_A} a$ $\overline \lambda \in \map {\sigma_A} {a^\ast}$.
We conclude:
:$\map {\sigma_A} {a^\ast} = \set {\overline \lambda : \lambda \in \map {\sigma_A} a}$
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:Spectra (Spectral Theory)
Spectrum of Unitary Element in Unital C*-Algebra is Subset of Unit Circle

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be unitary. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$. 
== Proof ==
Let $A' \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$.
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $A'$ is commutative. 
Let $\Phi_{A'}$ be the spectrum of $A'$.
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_{A'} } x$
From Spectrum of Element of Unital Commutative Banach Algebra, we have $\map {\sigma_A} x = \map {\sigma_{A'} } x = \set {\map \phi x : \phi \in \Phi_{A'} }$. 
From Character on Unital C*-Algebra has Modulus One at Unitary Elements, we have $\cmod {\map \phi x} = 1$ for each $\phi \in \Phi_{A'}$.
Hence $\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$. 
Category:C*-Algebras
Category:Unitary Elements of Unital *-Algebras
Spectrum of Zero Vector in Algebra

== Theorem ==
Let $A$ be an algebra over $\C$. 
Let ${\mathbf 0}_A$ be the zero vector of $A$.
Let $\map {\sigma_A} { {\mathbf 0}_A}$ be the spectrum of ${\mathbf 0}_A$ in $A$.
Then:
:$\map {\sigma_A} { {\mathbf 0}_A} = \set 0$ if $A \ne \set { {\mathbf 0}_A}$
:$\map {\sigma_A} { {\mathbf 0}_A} = \O$ if $A = \set { {\mathbf 0}_A}$.
== Proof ==
First suppose that $A$ is unital with identity element ${\mathbf 1}_A$.
For $\lambda \in \C \setminus \set 0$, we have:
:$\lambda^{-1} {\mathbf 1}_A \paren {\lambda {\mathbf 1}_A - {\mathbf 0}_A} = {\mathbf 1}_A = \paren {\lambda {\mathbf 1}_A - {\mathbf 0}_A} \lambda^{-1} {\mathbf 1}_A$
Hence $\lambda \not \in \map {\sigma_A} { {\mathbf 0}_A}$.
If $A = \set { {\mathbf 0}_A}$, then we have:
:${\mathbf 0}_A^2 = {\mathbf 0}_A = {\mathbf 1}_A$
so $0 \not \in \map {\sigma_A} { {\mathbf 0}_A}$, and we get $\map {\sigma_A} { {\mathbf 0}_A} = \O$ in this case. 
Now suppose that $A \ne \set { {\mathbf 0}_A}$, then ${\mathbf 1}_A \ne {\mathbf 0}_A$.
We then have $a {\mathbf 0}_A = {\mathbf 0}_A a = {\mathbf 0}_A \ne {\mathbf 1}_A$ for all $a \in A$.
Hence $0 \in \map {\sigma_A} { {\mathbf 0}_A}$.
Hence we get $\map {\sigma_A} { {\mathbf 0}_A} = \set 0$ in this case.
Now suppose that $A$ is not unital.
Let $A_+$ be the unitization of $A$.
Then we have:
:$\map {\sigma_A} { {\mathbf 0}_A} = \map {\sigma_{A_+} } {\tuple { {\mathbf 0}_A, 0} } \cup \set 0$
From the unital case we have:
:$\map {\sigma_{A_+} } {\tuple { {\mathbf 0}_A, 0} } = \set 0$
and hence:
:$\map {\sigma_A} { {\mathbf 0}_A} = \set 0$
Category:Spectra (Spectral Theory)
Speed of Body under Free Fall from Height

== Theorem ==
Let an object $B$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $B$ fall a distance $s$.
Then:
:$v = \sqrt {2 g s}$
where:
:$v$ is the speed of $B$ after having fallen a distance $s$
:$g$ is the Acceleration Due to Gravity at the height through which $B$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Mechanics
Category:Gravity
Category:Speed of Body under Free Fall from Height
Speed of Body under Free Fall from Height/Proof 1

== Theorem ==
== Proof ==
From Body under Constant Acceleration: Velocity after Distance:
:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf g \cdot \mathbf s$
All dot products are between pairs of parallel vectors.
Thus by Cosine Formula for Dot Product:
:$v^2 = u^2 + 2 g s$
Here the body falls from rest, so:
:$\mathbf u = \mathbf 0$
Thus:
:$v^2 = 2 g s$
and so:
:$v = \sqrt {2 g s}$
Category:Speed of Body under Free Fall from Height