avewright/proofwiki-math huggingface dataset:
Theorem: Topological Space with Sigma-Locally Finite K-Network has Sigma-Locally Finite Network

Let $T = \struct {S, \tau}$ be a topological space which has a $\sigma$-locally finite $k$-network.
Then $T$ has a $\sigma$-locally finite network.

Proof:
Follows directly from K-Network is Network.
{{qed}}
Theorem: Taylor Series of Lorentz Factor at Low Speeds

The Lorentz factor has the following Taylor polynomial around $v = 0$:
{{begin-eqn}}
{{eqn | l = \map \gamma v
      | r = \frac 1 {\sqrt {1 - \frac {v^2} {c^2} } }
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\frac {\paren {2 n - 1} !!} {2 n} } \paren {\frac v c }^{2 n}
}}
{{eqn | o = \approx
      | r = 1 + \frac 1 2 \paren {\frac v c }^2 + \frac 3 8 \paren {\frac v c }^4 + \frac 5 {16} \paren {\frac v c}^6  + \cdots
}}
{{end-eqn}}
where
:$v$ is speed
:$c$ is the speed of light
:$!!$ is a double factorial

Proof:
{{Recall|Lorentz Factor|Lorentz factor}}
{{:Definition:Lorentz Factor}}
Let $x = v^2 / c^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map \gamma x
      | r = \frac 1 {\sqrt {1 - x} }
}}
{{eqn | r = \paren {1 - x}^{-\frac 1 2}
}}
{{end-eqn}}
Note that $\map \gamma 0 = 1$.
Proof by induction:
For all $n \in \mathbb N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map {\gamma^{(n)} } x = \frac {\paren {2 n - 1} !!} {2^n} \paren {1 - x}^{-\frac {2 n + 1} 2}$
where $\map {\gamma^{\paren n} } x$ is the $n^{th}$ derivative of $\map \gamma x$.
Theorem: T4 and R0 Space is T3.5

Let $T = \struct {S, \tau}$ be a $T_4$ space which is also an $R_0$ space.
Then $T$ is also a $T_{3 \frac 1 2}$ space.

Proof:
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space}}
{{:Definition:T3.5 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a $T_4$ space and also an $R_0$ space
{{Recall|T4 Space|$T_4$ space}}
{{:Definition:T4 Space/Definition 1}}
{{Recall|R0 Space|$R_0$ space}}
{{:Definition:R0 Space}}
{{ProofWanted|Should be similar to Normal Space is T3.5, but in this case we cannot assume that $\set y$ is closed as $T$ is not necessarily $T_1$}}
Theorem: Topological Space with Dispersion Point is not Empty

Let $T = \struct {S, \tau}$ be a topological space which has a dispersion point.
Then $T$ is not the empty space.

Proof:
Let $T = \struct {S, \tau}$ have a dispersion point $p$.
Then $p \in S$ and so $S \ne \O$.
The result follows.
{{qed}}
Theorem: Partition Space is Extremally Disconnected

Let $T = \struct {S, \tau}$ be a partition space.
Then $T$ is extremally disconnected.

Proof:
{{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
Let $U \in \tau$ be an open set of $T$.
By Subset of Partition Space is Open iff Closed, $U$ is also closed in $T$.
From Closed Set equals its Closure, $U$ equals its closure.
Hence the result by definition of extremally disconnected.
{{qed}}
Theorem: Totally Pathwise Disconnected Space is T1

Let $T = \struct {S, \tau}$ be a topological space which is totally pathwise disconnected space.
Then $T$ is a $T_1$ space.

Proof:
Let $T = \struct {S, \tau}$ be a totally pathwise disconnected space.
By Total Pathwise Disconnectedness is Hereditary, every subspace of $T$ is also totally pathwise disconnected.
In particular, every doubleton subspace of $T$ is totally pathwise disconnected.
Let $a, b \in S$ be arbitrary distinct points.
From Doubleton Totally Pathwise Disconnected Space is Discrete, $\set a$ and $\set b$ are both closed in $T$.
As $a$ and $b$ are arbitrary, it follows that all singleton subsets of $S$ are closed in $T$.
Hence the result, by definition of $T_1$ space.
{{qed}}
Theorem: Total Pathwise Disconnectedness is Hereditary

Let $T = \struct {S, \tau}$ be a topological space which is totally pathwise disconnected space.
Let $H$ be a subspace of $T$.
Then $H$ is also a totally pathwise disconnected space.

Proof:
{{Recall|Totally Pathwise Disconnected Space|totally pathwise disconnected space}}
{{:Definition:Totally Pathwise Disconnected Space/Definition 1}}
Let $T = \struct {S, \tau}$ be totally pathwise disconnected.
{{AimForCont}} $H$ is a subspace of $T$ which is not totally pathwise disconnected.
Then by definition there exists a path component $K$ of $H$ which contains $2$ elements.
But then $K$ is also a path component of $T$.
But as $T$ is totally pathwise disconnected, there can be no such path component.
From that contradiction, it cannot be the case that $H$ is not totally pathwise disconnected.
Hence the result.
{{qed}}
Theorem: Doubleton Totally Pathwise Disconnected Space is Discrete

Let $T = \struct {S, \tau}$ be a topological space whose underlying set $S$ is a doubleton.
Let $T$ be totally pathwise disconnected space.
Then $T$ is a discrete space.

Proof:
{{WLOG}}, let $S = \set {0, 1}$.
Hence let $T = \struct {\set {0, 1}, \tau}$ be a totally pathwise disconnected space.
{{AimForCont}} $T$ is not a discrete space.
{{WLOG}}, let $\set 1$ not be an open set of $T$.
Let $f: \closedint 0 1 \to S$ be the mapping defined as:
:$\map f x = \begin {cases} 0 & : x < \dfrac 1 2 \\ \\ 1 & : x \ge \frac 1 2 \end {cases}$
Then $f$ is a continuous mapping which is not constant.
This contradicts our assertion that $T$ is a totally pathwise disconnected space.
Hence, by Proof by Contradiction, $T$ is a discrete space.
{{qed}}
Theorem: Pseudometrizable Space is Fully T4

A pseudometrizable space $T = \struct {S, \tau}$ is a fully $T_4$ space.

Proof:
{{Recall|Fully T4 Space|fully $T_4$ space}}
{{:Definition:Fully T4 Space}}
{{proof wanted}}
Theorem: Locally Pseudometrizable Space is First-Countable

Let $T = \struct {S, \tau}$ be a topological space which is locally pseudometrizable.
Then $T$ is first-countable.

Proof:
Let $\NN$ be a pseudometrizable neighborhood.
The open balls of rational radius within $\NN$ form a local basis.
Hence the result from the definition of first-countable.
{{qed}}
Theorem: K1-Space which is K1-Hausdorff is K3

Let $T = \struct {S, \tau}$ be a topological space which is both a $k_1$-space and a $k_1$-Hausdorff space.
Then $T$ is a $k_3$-space.

Proof:
{{Recall|K3-Space|$k_3$-space}}
{{:Definition:K3-Space}}
Let $T = \struct {S, \tau}$ be both a $k_1$-space and a $k_1$-Hausdorff space.
{{Recall|K1-Space|$k_1$-space}}
{{:Definition:K1-Space/Definition 1}}
{{Recall|K1-Hausdorff Space|$k_1$-Hausdorff space}}
{{:Definition:K1-Hausdorff Space/Definition 2}}
The result follows.
{{qed}}
Theorem: Weakly Lindelöf Space with Sigma-Locally Finite Basis is Second-Countable

Let $T = \struct {S, \tau}$ be a weakly Lindelöf topological space.
Let $\BB$ be a $\sigma$-locally finite basis of $T$.
Then $T$ is second-countable.

Proof:
{{Refactor}}
Theorem: Point in T1 Space is Intersection of Open Sets

Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is a $T_1$ space
{{iff}}
:for every point $x \in S$, the singleton $\set x$ is the intersection of every open set $U$ of $T$ such that $x \in U$:
::$\forall x \in S: \set x = \ds \bigcap_{x \mathop \in U} \set {U \in \tau}$

Proof:
{{Recall|T1 Space|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
For arbitrary $x \in S$, let $\CC_x$ be the intersection of every open set $U$ of $T$ such that $x \in U$:
:$\CC_x = \ds \bigcap_{x \mathop \in U} \set {U \in \tau}$
Theorem: Hermitian Conjugate is Adjoint

Let $\mathbf A$ be an $n \times m$ matrix over $\mathbb C$ that represents a linear transformation
:$\LL : V \to W$
where
:$V$ is $m$-dimensional
:$W$ is $n$-dimensional
in some orthonormal bases for $V$ and $W$.
Then, the adjoint $\LL^*$ is given by $\mathbf A^\dagger$, the Hermitian conjugate of $\mathbf A$.

Proof:
Let $\mathbf x \in \mathbb C^m$ and $\mathbf y \in \mathbb C^n$ be vectors representing elements of $V$ and $W$, respectively.
Then:
{{begin-eqn}}
{{eqn | l = \innerprod {\mathbf A \mathbf x} {\mathbf y}
      | r = \mathbf y^\dagger \paren {\mathbf A \mathbf x}
      | c = {{Defof|Complex Vector Inner Product}}
}}
{{eqn | r = \paren {\mathbf y^\dagger \mathbf A} \mathbf x
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf A^\dagger \mathbf y}^\dagger \mathbf x
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \innerprod {\mathbf x} {\mathbf A^\dagger \mathbf y}
      | c = {{Defof|Complex Vector Inner Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Normal Matrix is Normal Operator

A normal matrix is a normal operator.

Proof:
Let $\mathbf A$ be a normal matrix.
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf A^\dagger
      | r = \mathbf A^\dagger \mathbf A
      | c = {{Defof|Normal Matrix}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf A \mathbf A^*
      | r = \mathbf A^* \mathbf A
      | c = Hermitian Conjugate is Adjoint
}}
{{end-eqn}}
{{qed}}
Category:Normal Matrices
Category:Normal Operators
31tsqwi9n3pjswb6twkch7vx2jfkks8
Theorem: Hermitian Matrix is Normal

All Hermitian matrices are normal.

Proof:
Let $\mathbf T$ be a Hermitian matrix.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf T \mathbf T^\dagger
      | r = \mathbf T \mathbf T
      | c = {{Defof|Hermitian Matrix}}
}}
{{eqn | r = \mathbf T^\dagger \mathbf T
      | c = {{Defof|Hermitian Matrix}}
}}
{{end-eqn}}
Thus, $\mathbf T$ is normal.
{{qed}}
Category:Hermitian Matrices
Category:Normal Matrices
jyry67ln981q28rg8cnf3g0xybfgn49
Theorem: Anti-Hermitian Matrix is Normal

All anti-Hermitian matrices are normal.

Proof:
Let $\mathbf T$ be an anti-Hermitian matrix.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf T \mathbf T^\dagger
      | r = \mathbf T \paren {-\mathbf T}
      | c = {{Defof|Anti-Hermitian Matrix}}
}}
{{eqn | r = -\mathbf T \mathbf T
      | c = Matrix Multiplication is Homogeneous of Degree 1
}}
{{eqn | r = \mathbf T^\dagger \mathbf T
      | c = {{Defof|Anti-Hermitian Matrix}}
}}
{{end-eqn}}
Thus, $\mathbf T$ is normal.
{{qed}}
Category:Normal Matrices
Category:Anti-Hermitian Matrices
lwnplnyxprp0u69djb1p67y62gq4z5c
Theorem: Unitary Matrix is Normal

All unitary matrices are normal.

Proof:
Let $\mathbf U$ be a unitary matrix of order $n$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf U \mathbf U^\dagger
      | r = \mathbf U \mathbf U^{-1}
      | c = {{Defof|Unitary Matrix}}
}}
{{eqn | r = \mathbf I_n
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \mathbf U^{-1} \mathbf U
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \mathbf U^\dagger \mathbf U
      | c = {{Defof|Unitary Matrix}}
}}
{{end-eqn}}
Thus, $\mathbf U$ is normal.
{{qed}}
Category:Unitary Matrices
Category:Normal Matrices
g8eifc1p5ux5rc5pmb1iwzrx0rr41xb
Theorem: Eigenvalues of Unitary Matrix have Unit Modulus

Let $\mathbf U$ be a unitary matrix.
Let $\lambda$ be an eigenvalue of $\mathbf U$.
Then:
:$\size \lambda = 1$

Proof:
Let $\mathbf v$ be an eigenvector of $\mathbf U$ associated with $\lambda$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf U \mathbf v}^2
      | r = \innerprod {\mathbf U \mathbf v} {\mathbf U \mathbf v}
      | c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {\lambda \mathbf v} {\lambda \mathbf v}
      | c = {{Defof|Eigenvalue}} and {{Defof|Eigenvector}}
}}
{{eqn | r = \lambda \overline \lambda \innerprod {\mathbf v} {\mathbf v}
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \size \lambda^2 \innerprod {\mathbf v} {\mathbf v}
      | c = Modulus in Terms of Conjugate
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf U \mathbf v}^2
      | r = \innerprod {\mathbf U \mathbf v} {\mathbf U \mathbf v}
      | c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf U^* \mathbf U \mathbf v}
      | c = {{Defof|Adjoint Operator}}
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf U^\dagger \mathbf U \mathbf v}
      | c = Hermitian Conjugate is Adjoint
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf U^{-1} \mathbf U \mathbf v}
      | c = {{Defof|Unitary Matrix}}
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf v}
      | c = {{Defof|Inverse Matrix}}
}}
{{end-eqn}}
Therefore:
:$\size \lambda^2 \innerprod {\mathbf v} {\mathbf v} = \innerprod {\mathbf v} {\mathbf v}$
Since $\mathbf 0$ is never an eigenvector, $\innerprod {\mathbf v} {\mathbf v} > 0$ as required of an inner product.
So, we can divide by $\innerprod {\mathbf v} {\mathbf v}$ to obtain:
:$\size \lambda^2 = 1$
And thus:
:$\size \lambda = 1$
{{qed}}
Category:Unitary Matrices
Category:Eigenvalues of Square Matrices
g38d0hjedgqg6mavv1wdbeh0ae42n4c
Theorem: Weakly Sigma-Locally Compact Space is Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a weakly $\sigma$-locally compact space.
Then $T$ is weakly locally compact.

Proof:
{{Recall|Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space}}
{{:Definition:Weakly Sigma-Locally Compact Space}}
The result follows by definition.
{{qed}}
Theorem: Subcover of Open Cover is Open

Let $T = \struct {S, \tau}$ be a topological space.
Let $\CC$ be an open cover of $T$.
Let $\CC'$ be a subcover of $S$.
Then $\CC'$ is also an open cover of $T$.

Proof:
{{Recall|Open Cover|open cover}}
{{:Definition:Open Cover}}
{{Recall|Subcover|subcover}}
{{:Definition:Subcover}}
Let $U \in \CC'$ be arbitrary.
By definition of subcover, $U \in \CC$.
By definition of open cover, $U$ is open in $T$.
As $U$ is arbitrary, it follows that all elements of $\CC$ are open in $T$.
Hence, by definition, $\CC'$ is an open cover of $T$.
{{qed}}
Category:Open Covers
Category:Subcovers
1avfiicit0r2lzt8pxodbk9x7kjbhr1
Theorem: Product of Matrix and Standard Basis Vector is Column of Matrix

Let $\mathbb F$ be a field.
Let $\mathbf M$ be an $n \times m$ matrix over $\mathbb F$.
Let $\mathbf m_p$ be the $p^{th}$ column of $\mathbf M$.
Let $\mathbf e_p$ be the $p^{th}$ standard basis vector of $\mathbb F^m$ represented as a column vector.
Then:
:$\mathbf M \mathbf e_p = \mathbf m_p$

Proof:
By definition, since the order of $\mathbf M$ is $n \times m$ and the order of $\mathbf e_p$ is $m \times 1$, the order of $\mathbf M \mathbf e_p$ is $n \times 1$.
Thus, $\mathbf M \mathbf e_p$ is a column vector.
Let $\sqbrk {\mathbf M \mathbf e_p}_i$ be the $i^{th}$ component of $\mathbf M \mathbf e_p$.
Let $M_{ij}$ be an element of $\mathbf M$.
Let $e_{ip}$ be the $i^{th}$ component of $\mathbf e_p$.
We have:
{{begin-eqn}}
{{eqn | l = \sqbrk {\mathbf M \mathbf e_p}_i
      | r = \sum_{k \mathop = 1}^m M_{ik} e_{kp}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \sum_{k \mathop = 1}^m M_{ik} \delta_{kp}
      | c = {{Defof|Standard Ordered Basis on Vector Space}}
}}
{{eqn | r = M_{ip}
}}
{{end-eqn}}
The $i^{th}$ component of $\mathbf m_p$ is also $M_{ip}$.
Therefore,
:$\mathbf M \mathbf e_p = \mathbf m_p$
{{qed}}
Category:Conventional Matrix Multiplication
Category:Standard Ordered Bases
ky542b36uypbu35mtpgroqp2dqwwdy3
Theorem: Real Orthogonal Matrix is Unitary

All real orthogonal matrices are unitary.

Proof:
Let $\mathbf R$ be a real orthogonal matrix.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf R^\dagger
      | r = \mathbf R^\intercal
      | c = Hermitian Conjugate of Real Matrix is Transpose
}}
{{eqn | r = \mathbf R^{-1}
      | c = {{Defof|Orthogonal Matrix}}
}}
{{end-eqn}}
{{qed}}
Category:Orthogonal Matrices
Category:Unitary Matrices
62594wynzdtrj9wtm2652w2gl3aeztf
Theorem: Real Orthogonal Matrix is Normal

All real orthogonal matrices are normal.

Proof:
Follows directly from Real Orthogonal Matrix is Unitary and Unitary Matrix is Normal.
Category:Orthogonal Matrices
Category:Normal Matrices
jkm9uk6mt4vxvk5irw8c2dikrq10uo1
Theorem: Real Symmetric Matrix is Normal

All real symmetric matrices are normal.

Proof:
Follows directly from Real Symmetric Matrix is Hermitian and Hermitian Matrix is Normal.
Category:Real Matrices
Category:Symmetric Matrices
Category:Normal Matrices
2bnl4xd6ljue8ogkehtv45glhcqvye1
Theorem: Matrix Normality Preserved by Unitary Transformation

Let $\mathbf A$ and $\mathbf B$ be square matrices over $\mathbb C$.
Let $\mathbf A$ and $\mathbf B$ be similar via a unitary transformation, that is:
:$\mathbf B = \mathbf U \mathbf A \mathbf U^\dagger$
where
:$\mathbf U$ is a unitary matrix
:$\mathbf U^\dagger$ is the Hermitian conjugate of $\mathbf U$
Then, $\mathbf A$ is normal {{Iff}} $\mathbf B$ is normal.

Proof:
Let $\mathbf A$ be normal.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf B \mathbf B^\dagger
      | r = \paren {\mathbf U \mathbf A \mathbf U^\dagger} \paren {\mathbf U \mathbf A \mathbf U^\dagger}^\dagger
}}
{{eqn | r = \paren {\mathbf U \mathbf A \mathbf U^\dagger} \paren {\mathbf U \mathbf A^\dagger \mathbf U^\dagger}
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \mathbf U \mathbf A \paren {\mathbf U^\dagger \mathbf U} \mathbf A^\dagger \mathbf U^\dagger
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \mathbf U \mathbf A \mathbf A^\dagger \mathbf U^\dagger
      | c = $\mathbf U$ is unitary
}}
{{eqn | r = \mathbf U \mathbf A^\dagger \mathbf A \mathbf U^\dagger
      | c = $\mathbf A$ is normal
}}
{{eqn | r = \mathbf U \mathbf A^\dagger \paren {\mathbf U^\dagger \mathbf U} \mathbf A \mathbf U^\dagger
      | c = $\mathbf U$ is unitary
}}
{{eqn | r = \paren {\mathbf U \mathbf A^\dagger \mathbf U^\dagger} \paren {\mathbf U \mathbf A \mathbf U^\dagger}
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf U \mathbf A \mathbf U^\dagger}^\dagger \paren {\mathbf U \mathbf A \mathbf U^\dagger}
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \mathbf B^\dagger \mathbf B
}}
{{end-eqn}}
Therefore, $\mathbf B$ is normal.
{{qed|lemma}}
Let $\mathbf B$ be normal.
We have:
:$\mathbf B = \mathbf U \mathbf A \mathbf U^\dagger$
After pre-multiplying by $\mathbf U^\dagger$ and post-multiplying by $\mathbf U$, we have:
:$\mathbf A = \mathbf U^\dagger \mathbf B \mathbf U$
From Hermitian Conjugate of Unitary Matrix is Unitary, $\mathbf U^\dagger$ is unitary.
So, the proof that $\mathbf A$ is normal proceeds {{Mutatis}} as $\mathbf B$.
{{qed}}
Category:Matrix Similarity
Category:Normal Matrices
Category:Unitary Transformations
jbn8ek0cvcbmoisxn4cgj8ivj8rh38f
Theorem: Finite Cover is Locally Finite

Let $S$ be a set.
Let $\CC$ be a finite cover for a set $S$.
Then $\CC$ is also a locally finite cover for $S$.

Proof:
{{Recall|Locally Finite Cover|locally finite cover}}
{{:Definition:Locally Finite Cover}}
Let $\CC$ be a finite cover for a set $S$.
{{Recall|Finite Cover|finite cover}}
{{:Definition:Finite Cover}}
Let $x \in S$ be arbitrary.
From Topological Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$.
As $\CC$ is a finite cover, there are only a finite number of elements of $\CC$ that $S$ can intersect.
So $x$ has a neighborhood that intersects a finite number of elements of $\CC$.
As $x$ is arbitrary, it follows that every element of $S$ has a neighborhood that intersects a finite number of elements of $\CC$.
Hence the result by definition of locally finite cover.
{{qed}}
Theorem: Characterization of Diagonalizable Matrices

Let $\mathbf A$ be a diagonalizable matrix of order $n$ over a field $\mathbb F$:
:$\mathbf A = \mathbf X \mathbf D \mathbf X^{-1}$
where
{{begin-itemize}}
{{item||$\mathbf X$ is a nonsingular matrix of order $n$}}
{{item||$\mathbf D$ is a diagonal matrix of order $n$.}}
{{end-itemize}}
Then:
{{begin-itemize}}
{{item|(1):|the diagonal elements of $\mathbf D$ are eigenvalues of $\mathbf A$}}
{{item|(2):|the columns of $\mathbf X$ are the eigenvectors of $\mathbf A$ in the same order as their associated eigenvalues in $\mathbf D$.}}
{{end-itemize}}

Proof:
We have that $\mathbf D$ is diagonal.
Let $d_i = \sqbrk {\mathbf D}_{ii}$ be the $i^{th}$ diagonal element of $\mathbf D$.
Let $\mathbf x_i$ be a column of $\mathbf X$.
Let $\mathbf e_i$ be the $i^{th}$ standard basis vector of $\mathbb F^n$.
Therefore:
{{begin-eqn}}
{{eqn | l = \mathbf A
      | r = \mathbf X \mathbf D \mathbf X^{-1}
}}
{{eqn | ll= \leadsto
      | l = \mathbf A \mathbf X
      | r = \mathbf X \mathbf D
      | c = Post-multiply both sides by $\mathbf X$
}}
{{eqn | ll=\leadsto
      | l = \mathbf A \begin{bmatrix} \mathbf x_1 & \cdots & \mathbf x_n \end{bmatrix}
      | r = \mathbf X \begin{bmatrix} d_1 \mathbf e_1 & \cdots & d_n \mathbf e_n \end{bmatrix}
      | c = Write $\mathbf X$ and $\mathbf D$ in terms of their columns
}}
{{eqn | ll= \leadsto
      | l = \begin{bmatrix} \mathbf A \mathbf x_1 & \cdots & \mathbf A \mathbf x_n \end{bmatrix}
      | r = \begin{bmatrix} \mathbf X \paren {d_1 \mathbf e_1} & \cdots & \mathbf X \paren {d_n \mathbf e_n} \end{bmatrix}
      | c = {{Defof|Block Multiplication}}
}}
{{eqn | r = \begin{bmatrix} d_1 \mathbf X \mathbf e_1 & \cdots & d_n \mathbf X \mathbf e_n \end{bmatrix}
      | c = Matrix Multiplication is Homogeneous of Degree 1
}}
{{eqn | r = \begin{bmatrix} d_1 \mathbf x_1 & \cdots & d_n \mathbf x_n \end{bmatrix}
      | c = Product of Matrix and Standard Basis Vector is Column of Matrix
}}
{{end-eqn}}
Equal matrices must have equal columns:
:$\mathbf A \mathbf x_i = d_i \mathbf x_i$
Therefore, $d_i$, the diagonal element of $\mathbf D$, is an eigenvalue of $\mathbf A$.
Also, $\mathbf x_i$, the column of $\mathbf X$, is an eigenvector of $\mathbf A$ associated with $d_i$.
{{qed}}
Theorem: Upper Triangular Normal Matrix is Diagonal

A square matrix that is both upper triangular and normal is diagonal.

Proof:
Proof by induction.
For all $n \in \mathbb N_{>0}$, let $\map P n$ be the proposition:
:An $n \times n$ upper triangular normal matrix is diagonal
Theorem: Spectral Theorem for Normal Matrices

Let $\mathbf A$ be a square matrix over $\mathbb C$.
Then $\mathbf A$ is normal {{Iff}} it is diagonalizable by a unitary transformation.
That is, we can write:
:$\mathbf A = \mathbf U \mathbf D \mathbf U^\dagger$
where
:$\mathbf U$ is a unitary matrix
:$\mathbf D$ is a diagonal matrix

Proof:
=== Necessary Case ===
Assume $\mathbf A$ is diagonalizable by a unitary transformation.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf A^\dagger
      | r = \paren {\mathbf U \mathbf D \mathbf U^\dagger} \paren {\mathbf U \mathbf D \mathbf U^\dagger}^\dagger
}}
{{eqn | r = \paren {\mathbf U \mathbf D \mathbf U^\dagger} \paren {\mathbf U \mathbf D^\dagger \mathbf U^\dagger}
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \mathbf U \mathbf D \paren {\mathbf U^\dagger \mathbf U} \mathbf D^\dagger \mathbf U^\dagger
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \mathbf U \mathbf D \mathbf D^\dagger \mathbf U^\dagger
      | c = $\mathbf U^\dagger \mathbf U = \mathbf I$
}}
{{eqn | r = \mathbf U \mathbf D^\dagger \mathbf D \mathbf U^\dagger
      | c = Matrix Multiplication on Diagonal Matrices is Commutative
}}
{{eqn | r = \mathbf U \mathbf D^\dagger \paren {\mathbf U^\dagger \mathbf U} \mathbf D \mathbf U^\dagger
      | c = $\mathbf U^\dagger \mathbf U = \mathbf I$
}}
{{eqn | r = \paren {\mathbf U \mathbf D^\dagger \mathbf U^\dagger} \paren {\mathbf U \mathbf D \mathbf U^\dagger}
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf U \mathbf D \mathbf U^\dagger}^\dagger \paren {\mathbf U \mathbf D \mathbf U^\dagger}
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \mathbf A^\dagger \mathbf A
}}
{{end-eqn}}
Therefore, $\mathbf A$ is normal.
{{qed|lemma}}