avewright/proofwiki-math huggingface dataset:
Theorem: Adjoint is Conjugate Linear

Let $\HH$ and $\KK$ be Hilbert spaces over $\Bbb F \in \set {\R, \C}$.
Let $\map \BB {\HH, \KK}$ be the set of bounded linear transformations from $\HH$ to $\KK$.
Let $A, B \in \map \BB {\HH, \KK}$ be bounded linear transformations.
Let $\lambda \in \mathbb F$ be a scalar.
Then:
:$\paren {A + \lambda B}^* = A^* + \bar \lambda B^*$
where
:$A^*$ is the adjoint of $A$
:$\bar \lambda$ is the complex conjugate of $\lambda$
That is, the adjoint is conjugate linear.

Proof:
Let $x \in \HH$ and $y \in \KK$ be vectors.
We have:
{{begin-eqn}}
{{eqn | l = \innerprod x {\paren {A + \lambda B}^* y}
      | r = \innerprod {\paren {A + \lambda B} x} y
      | c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | r = \innerprod {A x + \lambda B x} y
}}
{{eqn | r = \innerprod {A x} y + \lambda \innerprod {B x} y
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \innerprod x {A^* y} + \lambda \innerprod x {B^* y}
      | c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | r = \innerprod x {A^* y} + \innerprod x {\bar \lambda B^* y}
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \innerprod x {A^* y + \bar \lambda B^* y}
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \innerprod x {\paren {A^* + \bar \lambda B^*} y}
}}
{{end-eqn}}
Therefore, by the Existence and Uniqueness of Adjoint:
:$\paren {A + \lambda B}^* = A^* + \bar \lambda B^*$
{{qed}}
Theorem: Adjoint is Conjugate Linear

Let $\HH$ and $\KK$ be Hilbert spaces over $\Bbb F \in \set {\R, \C}$.
Let $\map \BB {\HH, \KK}$ be the set of bounded linear transformations from $\HH$ to $\KK$.
Let $A, B \in \map \BB {\HH, \KK}$ be bounded linear transformations.
Let $\lambda \in \mathbb F$ be a scalar.
Then:
:$\paren {A + \lambda B}^* = A^* + \bar \lambda B^*$
where
:$A^*$ is the adjoint of $A$
:$\bar \lambda$ is the complex conjugate of $\lambda$
That is, the adjoint is conjugate linear.

Proof:
Let $\innerprod \cdot \cdot_\HH$ and $\innerprod \cdot \cdot_\KK$ be inner products on $\HH$ and $\KK$ respectively.
=== Lemma $1$ ===
{{:Adjoint is Conjugate Linear/Lemma 1}}{{qed|lemma}}
=== Lemma $2$ ===
{{:Adjoint is Conjugate Linear/Lemma 2}}{{qed|lemma}}
Thus:
{{begin-eqn}}
{{eqn | l = \paren {A + \lambda B}^*
      | r = A^* + \paren {\lambda B}^*
      | c = {{Lemma|Adjoint is Conjugate Linear|2}}
}}
{{eqn | r = A^* + \bar \lambda B^*
      | c = {{Lemma|Adjoint is Conjugate Linear|1}}
}}
{{end-eqn}}
{{qed}}
Theorem: Operator is Normal iff Operator Minus Multiple of Identity Operator is Normal

Let $\HH$ be a Hilbert space.
Let $\mathbf T: \HH \to \HH$ be a bounded linear operator.
Let $\lambda$ be a scalar under $\HH$.
Let $\mathbf I$ be the identity operator.
Then $\mathbf T$ is normal {{iff}} $\mathbf T - \lambda \mathbf I$ is normal.

Proof:
=== Sufficient Case ===
We have that $\mathbf T$ is normal.
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf T - \lambda \mathbf I} \paren {\mathbf T - \lambda \mathbf I}^*
      | r = \paren {\mathbf T - \lambda \mathbf I} \paren {\mathbf T^* - \overline \lambda \mathbf I^*}
      | c = Adjoint is Conjugate Linear
}}
{{eqn | r = \mathbf T \mathbf T^* - \overline \lambda \mathbf T \mathbf I^* - \lambda \mathbf I \mathbf T^* + \lambda \overline \lambda \mathbf I \mathbf I^*
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \mathbf T^* \mathbf T - \overline \lambda \mathbf T \mathbf I^* - \lambda \mathbf I \mathbf T^* + \lambda \overline \lambda \mathbf I^* \mathbf I
      | c = {{Defof|Normal Operator}} and Identity Mapping is Normal
}}
{{eqn | r = \mathbf T^* \mathbf T - \overline \lambda \mathbf I^* \mathbf T - \lambda \mathbf T^* \mathbf I + \lambda \overline \lambda \mathbf I^* \mathbf I
      | c = Identity Mapping is Hermitian: $\mathbf I^* \mathbf T = \mathbf I \mathbf T = \mathbf T = \mathbf T \mathbf I = \mathbf T \mathbf I^*$
}}
{{eqn | r = \paren {\mathbf T^* - \overline \lambda \mathbf I^*} \paren {\mathbf T - \lambda \mathbf I}
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \paren {\mathbf T - \lambda \mathbf I}^* \paren {\mathbf T - \lambda \mathbf I}
      | c = Adjoint is Conjugate Linear
}}
{{end-eqn}}
Thus, $\mathbf T - \lambda \mathbf I$ is normal.
{{qed|lemma}}
Theorem: Adjoint of Normal Operator has Same Eigenvectors and Complex Conjugated Eigenvalues

Let $\HH$ be a Hilbert space.
Let $\mathbf T: \HH \to \HH$ be a normal operator.
Let $\lambda$ be an eigenvalue of $\mathbf T$.
Let $\mathbf v$ be a eigenvector of $\mathbf T$ associated with $\lambda$.
Then, $\mathbf v$ is also an eigenvector of the adjoint $\mathbf T^*$ with an associated eigenvalue $\overline \lambda$, where $\bar \lambda$ is the complex conjugate of $\lambda$.
That is:
:$\mathbf T^* \mathbf v = \bar \lambda \mathbf v$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \mathbf {T v}
      | r = \lambda \mathbf v
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf 0
      | r = \mathbf {T v} - \lambda \mathbf v
}}
{{eqn | ll= \leadstoandfrom
      | l = 0
      | r = \norm {\paren {\mathbf T - \lambda \mathbf I} \mathbf v}
      | c = Norm is positive definite
}}
{{eqn | ll= \leadstoandfrom
      | l = 0
      | r = \norm {\paren {\mathbf T - \lambda \mathbf I}^* \mathbf v}
      | c = Operator is Normal iff Operator Minus Multiple of Identity Operator is Normal and Characterization of Normal Operators
}}
{{eqn | ll= \leadstoandfrom
      | l = 0
      | r = \norm {\mathbf T^* \mathbf v - \overline \lambda \mathbf v}
      | c = Adjoint is Conjugate Linear
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf 0
      | r = \mathbf T^* \mathbf v - \overline \lambda \mathbf v
      | c = Norm is positive definite
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf T^* \mathbf v
      | r = \overline \lambda \mathbf v
}}
{{end-eqn}}
Therefore, $\mathbf v$ is also an eigenvector of $\mathbf T^*$ with an associated eigenvalue $\bar \lambda$.
{{qed}}
Category:Normal Operators
Category:Eigenvectors of Linear Operators
Category:Eigenvalues of Linear Operators
Category:Adjoints
egzn06gwwufvx0mfkagfejlunuv0xz2
Theorem: Eigenvalues of Linear Operator Shifted by Scaled Identity Operator

Let $V$ be a vector space over a field $\mathbb F$.
Let $A : V \to V$ be a linear operator.
Let $s$ be a scalar in $\mathbb F$.
Let $I$ be the identity operator on $V$.
Then, the operator $T + s I$ has the same eigenvectors as $A$ with all eigenvalues shifted by $s$.

Proof:
Let $\lambda$ be an eigenvalue of $A$.
Let $v$ be an eigenvector of $A$ associated with $\lambda$.
We have:
{{begin-eqn}}
{{eqn | l = \paren {T + s I} v
      | r = T v + s v
}}
{{eqn | r = \lambda v + s v
}}
{{eqn | r = \paren {\lambda + s} v
}}
{{end-eqn}}
Therefore, $v$ is an eigenvector of $T + s I$ with a shifted eigenvalue of $\lambda + s$.
{{qed}}
Category:Linear Operators
Category:Eigenvalues of Linear Operators
Category:Identity Mappings
1gyoswttv4e98qiyk87o1og77k95qxk
Theorem: Eckmann-Hilton Argument

Let $S$ be a non-empty set.
Let $\circ$ and $\ast$ be binary operations on $S$ such that:
{{begin-itemize}}
{{item|(\text a):|$\circ$ and $\ast$ have respective identity elements $e_\circ$ and $e_\ast$ in $S$}}
{{item|(\text b):|$\forall a, b, c, d \in S: \paren {a \circ b} \ast \paren {c \circ d} {{=}} \paren {a \ast c} \circ \paren {b \ast d}$}}
{{end-itemize}}
Then:
{{begin-itemize}}
{{item|(1):|$e_\circ {{=}} e_\ast$}}
{{item|(2):|$\circ {{=}} \ast$}}
{{item|(3):|$\circ$ is commutative}}
{{end-itemize}}

Proof:
{{begin-eqn}}
{{eqn | l = e_\circ
      | r = e_\circ \circ e_\circ
      | c = {{Defof|Identity Element}}: $e_\circ$
}}
{{eqn | r = \paren {e_\circ \ast e_\ast} \circ \paren {e_\ast \ast e_\circ}
      | c = {{Defof|Identity Element}}: $e_\ast$
}}
{{eqn | r = \paren {e_\circ \circ e_\ast} \ast \paren {e_\ast \circ e_\circ}
      | c = from $(\text b)$
}}
{{eqn | r = e_\ast \ast e_\ast
      | c = {{Defof|Identity Element}}: $e_\circ$
}}
{{eqn | r = e_\ast
      | c = {{Defof|Identity Element}}: $e_\ast$
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = x \circ y
      | r = \paren {x \ast e_\ast} \circ \paren {e_\ast \ast y}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {x \circ e_\ast} \ast \paren {e_\ast \circ y}
      | c = from $(\text b)$
}}
{{eqn | r = \paren {x \circ e_\circ} \ast \paren {e_\circ \circ y}
      | c = from $(1)$
}}
{{eqn | r = x \ast y
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
From Equality of Mappings, $\circ = \ast$.
{{qed|lemma}}
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = x \circ y
      | r = \paren {e_\ast \ast x} \circ \paren {y \ast e_\ast}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {e_\ast \circ y} \ast \paren {x \circ e_\ast}
      | c = from $(\text b)$
}}
{{eqn | r = \paren {e_\ast \ast y} \circ \paren {x \ast e_\ast}
      | c = from $(2)$
}}
{{eqn | r = y \circ x
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
So $\circ$ is commutative.
{{qed}}
{{Namedfor|Beno Eckmann|name2 = Peter John Hilton|cat = Eckmann|cat2 = Hilton}}
Theorem: Peirce's Law/Strong Form/Formulation 1

:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$

Proof:
=== $(1):$ $\vdash$ Direction ===
{{:Peirce's Law/Strong Form/Formulation 1/Forward Direction}}
Theorem: Peirce's Law/Strong Form/Formulation 1

:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$

Proof:
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||c|}\hline
((p & \implies & q) & \implies & p) & p \\
\hline
\F & \T & \F & \F & \F & \F \\
\F & \T & \T & \F & \F & \F \\
\T & \F & \F & \T & \T & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
{{qed}}
Theorem: Peirce's Law/Strong Form/Formulation 1/Forward Direction

:$\paren {\paren {p \implies q} \implies p} \vdash p$

Proof:
This is none other than Peirce's law itself, whose proof we transclude here:
{{:Peirce's Law/Formulation 1/Proof 2}}
Theorem: Peirce's Law/Strong Form/Formulation 1/Reverse Direction

:$\paren {\paren {p \implies q} \implies p} \dashv p$

Proof:
{{BeginTableau|p \vdash \paren {p \implies q} \implies p}}
{{Premise|1|p}}
{{SequentIntro|2|1|\paren {p \implies q} \implies p|1|True Statement is implied by Every Statement}}
{{EndTableau|qed}}
Theorem: Peirce's Law/Strong Form/Formulation 2/Forward Direction

:$\paren {\paren {p \implies q} \implies p} \implies p$

Proof:
This is none other than Peirce's law itself, whose proof we transclude here:
{{:Peirce's Law/Formulation 2/Proof 1}}
Theorem: Peirce's Law/Strong Form/Formulation 2/Reverse Direction

:$\vdash p \implies \paren {\paren {p \implies q} \implies p}$

Proof:
{{BeginTableau|\vdash p \implies \paren {\paren {p \implies q} \implies p} }}
{{Assumption|1|p}}
{{SequentIntro|2|1|\paren {p \implies q} \implies p|1|Peirce's Law Strong Form: Formulation 1: $p \vdash \paren {p \implies q} \implies p$}}
{{Implication|3||p \implies \paren {\paren {p \implies q} \implies p}|1|2}}
{{EndTableau}}
{{qed}}
Theorem: (p implies q) implies q, q implies p therefore p

:$\paren {p \implies q} \implies q, q \implies p \vdash p$

Proof:
{{BeginTableau|\paren {p \implies q} \implies q, q \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies q}}
{{Premise|2|q \implies p}}
{{Assumption|3|\neg p}}
{{ModusTollens|4|2,3|\neg q|2|3}}
{{ModusTollens|5|1,2,3|\neg \paren {p \implies q}|1|4}}
{{SequentIntro|6|3|p \implies q|3|False Statement implies Every Statement}}
{{NonContradiction|7|1,2,3|5|6}}
{{Contradiction|8|1,2|p|3|7}}
{{EndTableau|qed}}
Theorem: (p implies q) implies q, q implies p therefore p

:$\paren {p \implies q} \implies q, q \implies p \vdash p$

Proof:
{{BeginTableau|\paren {p \implies q} \implies q, q \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies q}}
{{Premise|2|q \implies p}}
{{Assumption|3|\neg p}}
{{ModusTollens|4|2, 3|\neg q|2|3}}
{{ModusTollens|5|1, 2, 3|\neg \paren {p \implies q}|1|4}}
{{Assumption|6|p}}
{{NonContradiction|7|3, 6|3|6}}
{{Explosion|8|3, 6|q|7}}
{{Implication|9|3|p \implies q|6|8}}
{{NonContradiction|10|1, 2, 3|5|9}}
{{Contradiction|11|1, 2|p|3|9}}
{{EndTableau|qed}}
Theorem: Simple Spectrum Approximation of Diagonalizable Matrix/Property 1

:$\ds \lim_{t \mathop \to 0} \map {\mathbf A} t {{=}} \mathbf A$

Proof:
{{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0} \map {\mathbf A} t
      | r = \map {\mathbf A} 0
}}
{{eqn | r = \mathbf A + 0 \mathbf P \mathbf E \mathbf P^{-1}
}}
{{eqn | r = \mathbf A
}}
{{end-eqn}}
{{qed}}
Category:Simple Spectrum Approximation of Diagonalizable Matrix
8op46iqjpu9w5ns2yycw7hlh6uevfs9
Theorem: Simple Spectrum Approximation of Diagonalizable Matrix/Property 2

:$\map {\mathbf A} t$ is diagonalizable for all values of $t$

Proof:
Let $\mathbf D = \mathbf P^{-1} \mathbf A \mathbf P$ be $\mathbf A$ diagonalized.
{{begin-eqn}}
{{eqn | l = \map {\mathbf A} t
      | r = \mathbf A + t \mathbf P \mathbf E \mathbf P^{-1}
}}
{{eqn | r = \mathbf P \mathbf D \mathbf P^{-1} + t \mathbf P \mathbf E \mathbf P^{-1}
}}
{{eqn | r = \mathbf P \paren {\mathbf D + t \mathbf E} \mathbf P^{-1}
}}
{{end-eqn}}
Since $t \mathbf E$ is diagonal, $\mathbf D + t \mathbf E$ is diagonal.
By the definition of diagonalizable matrix, since $\map {\mathbf A} t$ is similar to a diagonal matrix $\mathbf D + t \mathbf E$, $\map {\mathbf A} t$ is diagonalizable for all values of $t$.
{{qed}}
Category:Simple Spectrum Approximation of Diagonalizable Matrix
mxh5jubaapvy77ftrn7259vci5g655v
Theorem: Simple Spectrum Approximation of Diagonalizable Matrix/Property 4

:$\ds \lim_{t \mathop \to 0} \map \sigma {\map {\mathbf A} t} {{=}} \map \sigma {\mathbf A}$

Proof:
From a previous proof, we have that:
:$\map {\mathbf A} t = \mathbf P \paren {\mathbf D + t \mathbf E} \mathbf P^{-1}$
where $\mathbf D = \mathbf P^{-1} \mathbf A \mathbf P$
Let $\varepsilon_i$ be the $i^{th}$ diagonal element of $\mathbf E$.
By Characterization of Diagonalizable Matrices, the eigenvalues of $\map {\mathbf A} t$ are the diagonal elements of $\mathbf D + t \mathbf E$, which are $\lambda_i + t \varepsilon_i$.
Thus, we have:
:$\map {\mathbf A} t \mathbf v_i = \paren {\lambda_i + t \varepsilon_i} \mathbf v_i$
Therefore, the spectrum of $\map {\mathbf A} t$ is given by:
:$\map \sigma {\map {\mathbf A} t} = \set {\lambda_i + t \varepsilon_i : \lambda_i \in \map \sigma {\mathbf A}}$
Thus:
{{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0} \map \sigma {\map {\mathbf A} t}
      | r = \lim_{t \mathop \to 0} \set {\lambda_i + t \varepsilon_i : \lambda_i \in \map \sigma {\mathbf A} }
}}
{{eqn | r = \set {\lambda_i : \lambda_i \in \map \sigma {\mathbf A} }
}}
{{eqn | r= \map \sigma {\mathbf A}
}}
{{end-eqn}}
{{qed}}
Category:Simple Spectrum Approximation of Diagonalizable Matrix
ahl6sjstq7ic5qrbxb3o9u1abvj4ips
Theorem: Simple Spectrum Approximation of Diagonalizable Matrix/Property 5

:There exists $\tau \in \R_{>0}$ such that $0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is simple

Proof:
Let $\lambda_i = \lambda_j$ be a non-simple eigenvalue of $\mathbf A$.
Let $\varepsilon_i$ be the $i^{th}$ diagonal element of $\mathbf E$.
Then, by a previous proof, the corresponding eigenvalues of $\map {\mathbf A} t$ are
:$\lambda_i + t \varepsilon_i$
:$\lambda_j + t \varepsilon_j$
The difference between these when $t \neq 0$ is:
{{begin-eqn}}
{{eqn | l = \lambda_j + t \varepsilon_j - \paren {\lambda_i + t \varepsilon_i}
      | r = t \paren {\varepsilon_j - \varepsilon_i}
      | c = $\lambda_j = \lambda_i$
}}
{{eqn | o = \neq
      | r = 0
      | c = $t \neq 0 \wedge \varepsilon_j \neq \varepsilon_i$
}}
{{end-eqn}}
So, what were non-simple eigenvalues are now simple when $t \neq 0$.
Let:
:$\delta = \min \set {\size {\lambda_j - \lambda_i} : i, j \in \closedint 1 n \wedge \lambda_i \neq \lambda_j}$
that is, the smallest absolute difference between distinct eigenvalues of $\mathbf A$.
Let $\varepsilon = \max \set {\size {\varepsilon_j - \varepsilon_i} : i, j \in \closedint 1 n}$.
that is, the largest absolute difference between the diagonal elements of $\mathbf E$.
Let:
:$\tau = \dfrac \delta \varepsilon$
Then, for $0 < t < \tau$ and $\lambda_i \neq \lambda_j$:
{{begin-eqn}}
{{eqn | l = \size {\paren {\lambda_i + t \varepsilon_i} - \paren {\lambda_j + t \varepsilon_j} }
      | r = \size {\paren {\lambda_i - \lambda_j} - t \paren {\varepsilon_j - \varepsilon_i} }
}}
{{eqn | o = \geq
      | r = \size {\delta - t \paren {\varepsilon_j - \varepsilon_i} }
}}
{{eqn | o = \geq
      | r = \size {\delta - t \varepsilon }
}}
{{eqn | o = >
      | r = 0
      | c = $t \varepsilon < \tau \varepsilon = \delta$
}}
{{end-eqn}}
Thus, distinct eigenvalues of $\mathbf A$ remain distinct.
Therefore, when $0 < t < \tau$, the spectrum of $\map {\mathbf A} t$ is simple.
{{qed}}
Category:Simple Spectrum Approximation of Diagonalizable Matrix
mf7ubky3ulx2fei4a9p80yf2nn57jke
Theorem: Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma

Let $\set {v_i}$ be a subset of eigenvectors of $A$ each associated with a distinct eigenvalue.
Then:
:$\ds \sum_i v_i \in W \implies \forall i : v_i \in W$

Proof:
Proof by induction.
Let $\map P n$ be the statement that:
:If $\size {\set {v_i}} = n$, then $\ds \sum_{i \mathop = 1}^n v_i \in W \implies \forall i \in \closedint 1 n : v_i \in W$
Theorem: Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable

Let $V$ be a vector space.
Let $A : V \to V$ be a diagonalizable linear operator.
Let $W \subseteq V$ be an invariant subspace of $\mathbf A$.
Let $A {\restriction_W}$ be the restriction of $A$ to $W$.
Then, $A {\restriction_W}$ is diagonalizable in $W$.

Proof:
=== Lemma ===
{{:Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma}}{{qed|lemma}}
Since $A$ is diagonalizable it has an eigenbasis.
Therefore, every vector in $W$ can be written as a linear combination of a subset of the eigenvectors of $A$.
By the {{Lemma|Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable}}, these eigenvectors are in $W$.
Therefore, $A {\restriction_W}$ has an eigenbasis of $W$.
Therefore, $A {\restriction_W}$ is diagonalizable.
{{qed}}
Category:Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable
Category:Diagonalizable Linear Operators
Category:Restrictions
6c04yagzh22fco9thdkl53tfq599nx8
Theorem: Diagonalizable Matrices Commute iff Simultaneously Diagonalizable

Let $\mathbb F$ be a field.
Let $M$ be a set of diagonalizable matrices over $\mathbb F$.
Then, $M$ is a commuting set {{Iff}} $M$ is simultaneously diagonalizable.

Proof:
=== Necessary Case ===
We have that $M$ is simultaneously diagonalizable.
Thus, there is a nonsingular matrix $\mathbf X$ that diagonalizes all elements of $M$.
Let $\mathbf A, \mathbf B \in M$ be arbitrary.
Let $\mathbf D_{\mathbf A} = \mathbf X^{-1} \mathbf A \mathbf X$
Let $\mathbf D_{\mathbf B} = \mathbf X^{-1} \mathbf B \mathbf X$
Note that $\mathbf D_{\mathbf A}$ and $\mathbf D_{\mathbf B}$ are diagonal.
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf B
      | r = \paren {\mathbf X^{-1} \mathbf D_{\mathbf A} \mathbf X} \paren {\mathbf X^{-1} \mathbf D_{\mathbf B} \mathbf X}
}}
{{eqn | r = \mathbf X^{-1} \mathbf D_{\mathbf A} \paren {\mathbf X \mathbf X^{-1} } \mathbf D_{\mathbf B} \mathbf X
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \mathbf X^{-1} \mathbf D_{\mathbf A} \mathbf D_{\mathbf B} \mathbf X
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \mathbf X^{-1} \mathbf D_{\mathbf B} \mathbf D_{\mathbf A} \mathbf X
      | c = Matrix Multiplication on Diagonal Matrices is Commutative
}}
{{eqn | r = \mathbf X^{-1} \mathbf D_{\mathbf B} \paren {\mathbf X \mathbf X^{-1} } \mathbf D_{\mathbf A} \mathbf X
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \paren {\mathbf X^{-1} \mathbf D_{\mathbf B} \mathbf X} \paren {\mathbf X^{-1} \mathbf D_{\mathbf A} \mathbf X}
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \mathbf B \mathbf A
}}
{{end-eqn}}
Therefore, $M$ is a commuting set.
{{qed|lemma}}
Theorem: Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication

:$\vdash \paren {p \land \paren {q \land r} } \impliedby \paren {\paren {p \land q} \land r}$

Proof:
{{BeginTableau|\vdash \paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} } }}
{{Assumption  |1|\paren {p \land q} \land r}}
{{Simplification|2|1|p \land q|1|1}}
{{Simplification|3|1|r|1|2}}
{{Simplification|4|1|p|2|1}}
{{Simplification|5|1|q|2|2}}
{{Conjunction|6|1|q \land r|5|3}}
{{Conjunction|7|1|p \land \paren {q \land r}|4|6}}
{{Implication |8||\paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} }|1|7}}
{{EndTableau|lemma}}
Theorem: Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication

:$\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}$

Proof:
{{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r} }}
{{Assumption  |1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land r|6|5}}
{{Implication |8||\paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}|1|7}}
{{EndTableau|lemma}}
Theorem: Rule of Commutation/Conjunction/Formulation 2/Proof 1/Forward Implication

:$\vdash \paren {p \land q} \implies \paren {q \land p}$

Proof:
{{BeginTableau|\vdash \paren {p \land q} \implies \paren {q \land p} }}
{{Assumption  |1|\paren {p \land q} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q|1|2}}
{{Conjunction|4|1|\paren {q \land p}|3|2}}
{{Implication|5||\paren {p \land q} \implies \paren {q \land p}|1|4}}
{{EndTableau|lemma}}
Theorem: Rule of Commutation/Conjunction/Formulation 2/Proof 1/Reverse Implication

:$\vdash \paren {q \land p} \implies \paren {p \land q}$

Proof:
{{BeginTableau|\vdash \paren {q \land p} \implies \paren {p \land q} }}
{{Assumption  |1|\paren {q \land p} }}
{{Simplification|2|1|q|1|1}}
{{Simplification|3|1|p|1|2}}
{{Conjunction|4|1|\paren {p \land q}|3|2}}
{{Implication|5||\paren {q \land p} \implies \paren {p \land q}|1|4}}
{{EndTableau|lemma}}
Theorem: Characteristic Polynomial of Matrix is Monic

Let $R$ be a commutative ring with unity.
Let $\mathbf A$ be a square matrix over $R$ of order $n > 0$.
Let $\mathbf I_n$ be the $n \times n$ identity matrix.
Let $R \sqbrk x$ be the polynomial ring in one variable over $R$.
Let $\map {p_{\mathbf A} } x$ be the characteristic polynomial of $\mathbf A$.
Then $\map {p_{\mathbf A} } x$ is monic.

Proof:
From Existence of Schur Decomposition for Square Matrix, matrix $\mathbf A$ is similar to an upper triangular matrix $\mathbf T$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {p_{\mathbf A} } x
      | r = \map \det {x \mathbf I_n - \mathbf A}
      | c = {{Defof|Characteristic Polynomial of Matrix}}
}}
{{eqn | r = \map \det {x \mathbf I_n - \mathbf T}
      | c = Similar Matrices have Same Characteristic Polynomial
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {x - T_{i i} }
      | c = Determinant of Upper Triangular Matrix
}}
{{end-eqn}}
where $T_{i i}$ is the $i^{th}$ diagonal element of $T$.
Each multiplicand $x - T_{i i}$ of the product is a monic polynomial because the coefficient of the highest power of $x$ is $1$.
From Product of Monic Polynomials, the product of $n$ monic polynomials is monic.
{{qed}}
Theorem: Characteristic Polynomial of Triangular Matrix

Let $\mathbf T$ be a square triangular matrix of order $n$.
Then, the characteristic polynomial $\map {p_T} x$ is given by:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$ are the diagonal elements of $\mathbf T$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map {p_T} x
      | r = \map \det {x \mathbf I - \mathbf T}
      | c = {{Defof|Characteristic Polynomial of Matrix}}
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }
      | c = Determinant of Triangular Matrix
}}
{{end-eqn}}
{{qed}}
Category:Triangular Matrices
Category:Characteristic Polynomial of Matrix
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