avewright/proofwiki-math huggingface dataset:
Theorem: Included Set Topology on Finite Intersection

Let $T = \struct {S, \tau}$ be a topological space on a set $S$.
Let $A_1, A_2, \ldots, A_n$ be a finite set of subsets of $S$:
:$\forall i \in \closedint 1 n: A_i \subseteq S$
Let $\forall i \in \closedint 1 n: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.
Let:
:$\forall i \in \closedint 1 n: \map T {A_i} \le T$
where $\map T {A_i} \le T$ denotes that $\map T {A_i}$ is coarser than $T$.
Then:
:$\ds \map T {\bigcap A_i} \le T$
where $\ds \map T {\bigcap A_i}$ is the included set space on $S$ by $\ds \bigcap_{i \mathop = 1}^n A_i$.

Proof:
For ease of notation, define:
:$A := \ds \bigcap_{i \mathop = 1}^n A_i$
and let $\tau_A$ denote the included set topology on $S$ by $A$.
Let $U \in \tau_A$ be nonempty.
Then by definition, $A \subseteq U$.
Hence there is a subset $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is:
:$U = \ds \paren {\bigcap_{i \mathop = 1}^n A_i} \cup Z = \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$
where the latter equality follows from Union Distributes over Intersection.
For every $i \in \closedint 1 n$, one has $A_i \cup Z \in \tau_{A_i}$ by definition of included set topology.
Further, by assumption, $\map T {A_i} \le T$, that is $\tau_{A_i} \subseteq \tau$.
It follows that $A_i \cup Z \in \tau$ for all $i \in \closedint 1 n$.
Hence $U = \ds \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ is in $\tau$, because $\tau$ is a topology.
This comes down to:
:$\tau_A \subseteq \tau$
Hence by definition of coarser topology:
:$\map T {\bigcap A_i} \le T$
{{qed}}
Theorem: Transformation of Unit Matrix into Inverse

Let $\mathbf A$ be a square matrix of order $n$ of the matrix space $\map {\MM_\R} n$.
Let $\mathbf I$ be the unit matrix of order $n$.
Suppose there exists a sequence of elementary row operations that reduces $\mathbf A$ to $\mathbf I$.
Then $\mathbf A$ is nonsingular.
Futhermore, the same sequence, when performed on $\mathbf I$, results in the inverse of $\mathbf A$.

Proof:
For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as:
{{begin-eqn}}
{{eqn | l = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf A
      | r = \mathbf I
}}
{{end-eqn}}
From Elementary Row Matrix is Nonsingular:
:$\mathbf E_1, \dotsc, \mathbf E_t \in \GL {n, \R}$
{{MissingLinks|$\GL {n, \R}$, and explain the significance of this. It's General Linear Group, clearly.}}
We can multiply on the left both sides of this equation by:
{{begin-eqn}}
{{eqn | l = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf  A 
      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
}}
{{eqn | ll= \leadsto
      | l = \mathbf {I I} \cdots \mathbf {I I A} 
      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf A
      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t
      | c = {{Defof|Unit Matrix}}
}}
{{eqn | ll= \leadsto
      | l = \breve {\mathbf A}
      | r = \breve {\breve {\mathbf E} }_t \breve {\breve {\mathbf E} }_{t - 1} \cdots \breve {\breve {\mathbf E} }_2 \breve {\breve {\mathbf E} }_1
      | c = Inverse of Matrix Product, Leibniz's Law
}}
{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1
      | c = Inverse of Group Inverse
}}
{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf I
      | c = {{Defof|Unit Matrix}}
}}
{{end-eqn}}
By repeated application of Elementary Row Operations as Matrix Multiplications, each $\mathbf E_n$ on the {{RHS}} corresponds to an elementary row operation.
Hence the result.
{{qed}}
{{proofread}}
Category:Unit Matrices
Category:Inverse Matrices
Category:Elementary Row Operations
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Theorem: Included Set Topology on Union

Let $T = \struct {S, \tau}$ be a topological space on a set $S$.
Let $\family {A_i}_{i \mathop \in I}$ be a family of subsets of $S$ indexed by the indexing set $I$:
:$\forall i \in I: A_i \subseteq S$
Let $\forall i \in I: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.
Let:
:$\forall i \in I: \map T {A_i} \ge T$
where $\map T {A_i} \ge T$ denotes that $\map T {A_i}$ is finer than $T$.
Then:
:$\ds \map T {\bigcup A_i} \ge T$
where $\ds \map T {\bigcup A_i}$ is the included set space on $S$ by $\ds \bigcup_{i \mathop \in I} A_i$.

Proof:
For ease of notation, define:
:$A := \ds \bigcup_{i \mathop \in I} A_i$
and let $\tau_A$ denote the included set topology on $S$ by $A$.
Let $U \in \tau$ be nonempty.
As $\map T {A_i}$ is finer than $T$ it follows by definition that:
:$\forall i \in I: \tau \subseteq \tau_{A_i}$
Thus:
:$\forall i \in I: U \in \tau_{A_i}$
Hence for all $i$ there is a subset $Z_i \subseteq S$ of $S$, such that $U = A_i \cup Z_i$; that is:
:$\ds U = \bigcup_{i \mathop \in I} \paren {A_i \cup Z_i} = \paren {\bigcup_{i \mathop \in I} A_i} \cup \paren {\bigcup_{i \mathop \in I} Z_i}$
where the latter equality follows from associativity and commutativity of set union.
That is:
:$U = A \cup Z$
where:
:$\ds Z = \paren {\bigcup_{i \mathop \in I} Z_i}$
Hence $U \in \tau_A$. by definition of the included set topology.
This comes down to $\tau \subseteq \tau_A$, and hence $\ds \map T {\bigcup A_i} \ge T$, by definition of a finer topology.
{{qed}}
Theorem: Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace

Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \map {\sigma_p} T, \lambda \ne 0$ be a nonzero eigenvalue of $T$.
Then the eigenspace for $\lambda$ has finite dimension.

Proof:
{{tidy}}
{{MissingLinks}}
$\def \sequence#1{\left({#1}\right)}$
Note that in the following, the notation for a sequence of terms in the form $e_n$ has been changed for this particular page to be $\sequence {e_n}$ rather than the {{ProofWiki}} standard $\left\langle{e_n}\right\rangle$.
This is in order to avoid potential confusion with the notation $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} }$ for inner product.
Assume that there is an infinite orthonormal sequence $\sequence {e_j}$ in my eigenspace for the eigenvalue $\lambda$.
That is:
:$\sequence {e_j: j \in J} \subset \map \ker {T - \lambda}$
Since $\norm {e_j} = 1$, then $\sequence {T e_n}$ will be in the closure image of $T$ of the closed unit ball.
Since the closure of this image is compact by assumption, then there is a subsequence $\sequence {T e_{n_k} }$ that converges, hence $\sequence {T e_{n_k} }$ is Cauchy.
Notice that
:$\norm {T e_{n_k} - Te_{n_p} }^2 = \norm {\lambda e_{n_k} - \lambda e_{n_p} }^2 = 2 \size \lambda^2 > 0$
for $k \ne p$, we are using the fact that $\lambda \ne 0$ and $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} } = 0$ (since $\sequence {e_j}$ is an orthonormal sequence).
Since $\lambda$ is constant, then it contradicts the fact that $\sequence {e_{n_k} }$ is Cauchy.
Thus $\map \ker {T - \lambda}$ is finite-dimensional.
{{qed}}
Theorem: Condition for Nonzero Eigenvalue of Compact Operator

Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero scalar.
Suppose that the following holds:
:$\inf \set {\norm {\paren {T - \lambda I} h}_H: \norm h_H = 1} = 0$
Then $\lambda \in \map {\sigma_p} T$, that is, $\lambda$ is an eigenvalue for $T$.

Proof:
{{proof wanted}}
By hypothesis, there is a sequence of unit vectors $\sequence{h_n}$ such that $\cmod{(T-\lambda) h_n} \to 0$.
Since $T$ is compact, by definition 2 there is a $f\in H$ and a subsequence $\sequence{h_{n_k}}$ such that $T h_{n_k}\to f$. 
But $h_{n_k}=\lambda^{-1}\left[(\lambda-T) h_{n_k}+T h_{n_k}\right] \to\lambda^{-1} f$.
So $1=\norm{\lambda^{-1} f}=\cmod\lambda^{-1}\norm{f}$ and $f \neq 0$.
Also, it must be that $T h_{n_k} \to \lambda^{-1} T f$.
Since $T h_{n_k} \to f, f=\lambda^{-1} T f$, or $T f=\lambda f$.
That is, $f \in \ker(T-\lambda)$ and $f \neq 0$, so $\lambda \in \sigma_p(T)$.
Theorem: Finite Rank Operator is Compact

Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_{00} } {H, K}$ be a bounded finite rank operator.
Then $T \in \map {B_0} {H, K}$, that is, $T$ is compact.

Proof:
{{tidy}}
{{MissingLinks}}
Let $A$ be a bounded set in $H$.
Since $T$ is bounded, the image $T \sqbrk A$ is a bounded set by definition.
Since $T$ is finite rank operator, its range $\Rng T$ is finite dimensional.
The closure of a bounded set in $\Rng T$ is compact, by Heine-Borel Theorem.
So the closure of $T \sqbrk A$ in $\Rng T$ is compact.
Hence, $T$ is a compact linear operator.
{{qed}}
Theorem: Adjoint of Finite Rank Operator

Let $\GF \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.
Then:
:$T^* \in \map {B_{00} } {\KK, \HH}$
that is, the adjoint of $T$ is also a bounded finite rank operator.

Proof:
From Characterization of Finite Rank Operators, there exists $e_1, e_2, \ldots, e_n \in \HH$ and $g_1, g_2, \ldots, g_n \in \KK$ such that: 
:$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$ 
for each $x \in \HH$. 
Then for each $x \in \HH$, $y \in \KK$ we have: 
{{begin-eqn}}
{{eqn	| l = \innerprod {T x} y_\KK
		| r = \innerprod {\sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i} y_\KK
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \innerprod {\innerprod x {e_i} g_i} y_\KK
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH \innerprod {g_i} y_\KK
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \innerprod x {e_i \overline {\innerprod {g_i} y_\KK} }_\HH
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \innerprod x {e_i \innerprod y {g_i}_\KK}_\HH
		| c = since the inner product is conjugate symmetric
}}
{{eqn	| r = \innerprod x {\sum_{i \mathop = 1}^n \innerprod y {g_i}_\KK e_i}_\HH
		| c = Inner Product is Sesquilinear
}}
{{end-eqn}}
From Existence and Uniqueness of Adjoint: Lemma 1, we therefore have: 
:$\ds T^\ast y = \sum_{i \mathop = 1}^n \innerprod y {g_i}_\KK e_i$
for each $y \in \KK$. 
So that: 
:$\ds T^\ast y \in \span \set {e_1, e_2, \ldots, e_n}$ 
for each $y \in \KK$. 
So we have that $T^\ast$ has finite rank.
{{qed}}
Theorem: Compact Idempotent is of Finite Rank

Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact linear operator.
Let $T$ be idempotent.
Then:
:$T \in \map {B_{00} } H$
that is, $T$ is a bounded finite rank operator.

Proof:
{{MissingLinks}}
{{tidy}}
Let $\sequence {y_n}$ be a bounded sequence in the range of $T$. 
Since $T$ is idempotent:
:$\forall n \in \N: T y_n = y_n$
Since $T$ is compact, $\sequence {T y_n}$ and hence $\sequence {y_n}$, contains a convergent subsequence.
Thus the range of $T$ has the Bolzano-Weierstrass property.
By Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition, the range has finite dimension.
{{qed}}
Theorem: Characterization of Finite Rank Operators

Let $\GF \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.
Let $n = \map \dim {\operatorname {ran} T}$ be the rank of $T$.
Then there are orthonormal vectors $e_1, \ldots, e_n \in \HH$ and vectors $g_1, \ldots, g_n \in \KK$ such that:
:$\forall h \in \HH: T h = \ds \sum_{i \mathop = 1}^n \innerprod h {e_i}_\HH g_i$

Proof:
Let: 
:$\set {e_1, e_2, \ldots, e_n}$
be a basis of $T \sqbrk \HH$.
By Gram-Schmidt Orthogonalization, there exists an orthonormal subset such that: 
:$\set {g_1, g_2, \ldots, g_n}$ 
that is a basis of $T \sqbrk \HH$.
So we have $T x \in \map \span {g_1, g_2, \ldots, g_n}$ for each $x \in \HH$.
So, we can write: 
:$\ds T x = \sum_{i \mathop = 1}^n \map {\alpha_i} x g_i$
for functions $\alpha_i : \HH \to \GF$. 
We show that each $\alpha_i$ is a bounded linear functional. 
For each $\lambda, \mu \in \GF$ and $x, y \in \HH$ such that: 
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \map {\alpha_i} {\lambda x + \mu y} g_i
		| r = \map T {\lambda x + \mu y}
}}
{{eqn	| r = \lambda T x + \mu T y
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} g_i
}}
{{end-eqn}}
So:
:$\ds \sum_{i \mathop = 1}^n \paren {\map {\alpha_i} {\lambda x + \mu y} - \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} } g_i = \mathbf 0_\KK$
From Orthogonal Set is Linearly Independent Set, $g_1, g_2, \ldots, g_n \in \KK$ are linearly independent. 
So we have: 
:$\map {\alpha_i} {\lambda x + \mu y} = \lambda \map {\alpha_i} x + \mu \map {\alpha_i} y$
for each $i$. 
So each $\alpha_i$ is a linear functional. 
From Pythagoras's Theorem (Inner Product Space), we have: 
:$\ds \norm {T x}^2_\KK = \sum_{i \mathop = 1}^n \cmod {\map {\alpha_i} x}^2$ 
for each $x \in \HH$. 
So we have: 
:$\cmod {\map {\alpha_i} x} \le \norm {T x}_\KK$
for each $x \in \HH$.
From Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\cmod {\map {\alpha_i} x} \le \norm T_{\map \BB {\HH, \KK} } \norm x_\HH$
for each $x \in \HH$. 
So each $\alpha_i$ is a bounded linear functional.
From Riesz Representation Theorem (Hilbert Spaces), there exists $e_i$ such that: 
:$\map {\alpha_i} x = \innerprod x {e_i}$
for each $x \in \HH$. 
Then we have: 
:$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$
for each $x \in \HH$. 
{{ProofWanted|Forgot to prove that the $e_i$s can be made orthonormal. This is easy, but will do later}}
{{qed}}
Theorem: Compact Hermitian Operator has Countable Point Spectrum

Let $\HH$ be a Hilbert space.
Let $T \in \map {B_0} \HH$ be a compact Hermitian operator.
{{explain|What is $\map {B_0} \HH$?}}
Then its point spectrum $\map {\sigma_p} T$ is countable.

Proof:
{{proof wanted|Use the next result, the spectral theorem}}
Theorem: Eigenspace for Normal Operator is Reducing Subspace

Let $\HH$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $A \in \map B \HH$ be a normal operator.
Let $\lambda \in \Bbb F$.
Let $I \in \map B \HH$ be the identity operator.
Then $\map \ker {A - \lambda I}$ is a reducing subspace for $A$.
Here $\ker$ denotes kernel.

Proof:
We are given that $A$ is normal.
Hence by {{Corollary|Kernel of Linear Transformation is Orthocomplement of Image of Adjoint}}:
:$\ker A = \Rng A^\perp$
and in particular, that:
:$\ker A \subseteq \Rng A^\perp$
Now, by Orthocomplement of Subset of Orthocomplement is Superset:
:$\Rng A \subseteq \ker A^\perp$
From Operator is Normal iff Operator Minus Multiple of Identity Operator is Normal, $A - \lambda I$ is normal.
Again by Orthocomplement of Subset of Orthocomplement is Superset, we find:
:$\Rng {A - \lambda I} \subseteq \paren {\map \ker {A - \lambda I} }^\perp$
We are now set up to prove that $\map \ker {A - \lambda I}$ is a reducing subspace for $A$.
Let $x \in \map \ker {A - \lambda I}$.
Then:
{{begin-eqn}}
{{eqn | l = A x
      | r = \lambda x + \paren {A - \lambda I} x
}}
{{eqn | r = \lambda x
      | c = {{Defof|Kernel of Linear Transformation}}, $x \in \map \ker {A - \lambda}$
}}
{{end-eqn}}
Therefore:
:$A \map \ker {A - \lambda I} \subseteq \map \ker {A - \lambda I}$
that is to say:
:$\map \ker {A - \lambda I}$ is $A$-invariant.
Now, let $x \in \paren {\map \ker {A - \lambda I} }^\perp$.
Observe that:
:$A x = \lambda x + \paren {A - \lambda I} x$
Now $\paren {A - \lambda I} x \in \Rng {A - \lambda I}$, and by our derivation above, this means that:
:$\paren {A - \lambda I} x \in \paren {\map \ker {A - \lambda I} }^\perp$
In conclusion, since $\paren {\map \ker {A - \lambda I} }^\perp$ is a linear subspace of $H$, it follows that:
:$\lambda x + \paren {A - \lambda I} x \in \paren {\map \ker {A - \lambda I} }^\perp$
as desired.
Hence both $\map \ker {A - \lambda I}$ and $\paren {\map \ker {A - \lambda I} }^\perp$ have been shown to be $A$-invariant subspaces of $H$.
That is, $\map \ker {A - \lambda I}$ is a reducing subspace for $A$.
{{qed}}
Theorem: Eigenvalues of Normal Operator have Orthogonal Eigenspaces

Let $\HH$ be a Hilbert space.
Let $\mathbf T: \HH \to \HH$ be a normal operator.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf T$.
Then, the eigenspaces of $\lambda_1$ and $\lambda_2$ are orthogonal:
:$\map \ker {\mathbf T - \lambda_1 \mathbf I} \perp \map \ker {\mathbf T - \lambda_2 \mathbf I}$
where:
:$\ker$ denotes kernel
:$\mathbf I$ denotes the identity operator
:$\perp$ denotes orthogonality.

Proof:
Let $\mathbf v_1$ and $\mathbf v_2$ be eigenvectors of $\mathbf T$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, respectively.
Then:
{{begin-eqn}}
{{eqn | l = \lambda_1 \innerprod {\mathbf v_1} {\mathbf v_2}
      | r = \innerprod {\lambda_1 \mathbf v_1} {\mathbf v_2}
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \innerprod {\mathbf T \mathbf v_1} {\mathbf v_2}
      | c = {{Defof|Eigenvector of Linear Operator}}
}}
{{eqn | r = \innerprod {\mathbf v_1} {\mathbf T^* \mathbf v_2}
      | c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | r = \innerprod {\mathbf v_1} {\overline{\lambda_2} \mathbf v_2}
      | c = Adjoint of Normal Operator has Same Eigenvectors and Complex Conjugated Eigenvalues
}}
{{eqn | r = \lambda_2 \innerprod {\mathbf v_1} {\mathbf v_2}
      | c = Inner Product is Sesquilinear
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = \paren {\lambda_1 - \lambda_2} \innerprod {\mathbf v_1} {\mathbf v_2}
}}
{{end-eqn}}
Since $\lambda_1 \ne \lambda_2$, this is only possible if $\innerprod {\mathbf v_1} {\mathbf v_2} = 0$, which means the eigenvectors of our normal operator are orthogonal.
{{qed}}
Theorem: Eigenvalues of Hermitian Operator are Real

Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a Hermitian operator.
Then all eigenvalues of $A$ are real.

Proof:
Let $\lambda$ be an eigenvalue of $A$.
Let $v \in \HH$ be an eigenvector for $\lambda$.
That is:
:$A v = \lambda v$
Now compute:
{{begin-eqn}}
{{eqn | l = \lambda \innerprod v v
      | r = \innerprod {\lambda v} v
      | c = Property $(2)$ of an inner product
}}
{{eqn | r = \innerprod {A v} v
      | c = $v$ is an eigenvector
}}
{{eqn | r = \innerprod v {A v}
      | c = $A$ is Hermitian
}}
{{eqn | r = \overline {\innerprod {A v} v}
      | c = Property $(1)$ of an inner product
}}
{{eqn | r = \overline {\innerprod {\lambda v} v}
      | c = $v$ is an eigenvector
}}
{{eqn | r = \bar \lambda \innerprod v v
      | c = Properties $(2)$, $(3)$ of an inner product 
}}
{{end-eqn}}
Now $v$, being an eigenvector, is non-zero.
By property $(4)$ of an inner product, this implies $\innerprod v v \ne 0$.
Thence, dividing out $\innerprod v v$, obtain $\lambda = \bar \lambda$.
From Complex Number equals Conjugate iff Wholly Real, $\lambda \in \R$.
{{qed}}
Theorem: Eigenvalues of Hermitian Operator are Real

Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a Hermitian operator.
Then all eigenvalues of $A$ are real.

Proof:
We use Dirac notation in the following:
Let $\ket x \in \HH$ be an eigenvector of $A$.
Let $\lambda \in \C$ be the associated eigenvalue:
That is:
:$A \ket x = \lambda \ket x$
For a general operator $T$ in $\map B \HH$ and $\ket x, \ket y \in \HH$:
{{begin-eqn}}
{{eqn | l = \bra x T \ket y
      | r = \braket x {T y}
      | c = {{Defof|Dirac Notation/Inner Product}}
}}
{{eqn | r = \overline {\braket {T y} x}
      | c = Property $(1)$ of Complex Inner Product
}}
{{eqn | r = \overline {\braket y {T^* x} }
      | c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | r = \overline {\bra y T^* \ket x}
      | c = {{Defof|Dirac Notation/Inner Product}}
}}
{{end-eqn}}
where the overline denotes the complex conjugate and $T^*$ denotes the adjoint of $T$.
Setting $T = A$ and noting that $A = A^*$ gives:
:$\bra x A \ket y = \overline {\bra y A \ket x}$
Now we compute:
{{begin-eqn}}
{{eqn | l = \bra x A \ket x
      | r = \braket x {A x}
      | c = 
}}
{{eqn | r = \braket x {\lambda x}
      | c = 
}}
{{eqn | r = \lambda \braket x x
      | c = In Dirac notation, inner products are linear in the second argument.
}}
{{end-eqn}}
Using our previous result:
{{begin-eqn}}
{{eqn | l = \bra x A \ket x
      | r = \overline {\bra x A \ket x}
      | c = 
}}
{{eqn | r = \overline {\lambda \braket x x}
      | c = 
}}
{{end-eqn}}
Equating the two previous equations gives:
:$\lambda \braket x x = \overline {\lambda \braket x x}$
Recalling the non-negative definiteness property of the inner product, we can see that:
:$\braket x x = \overline {\braket x x}$
So:
:$\lambda \braket x x = \overline \lambda \braket x x$
and so:
:$\lambda = \overline \lambda$
Therefore, by Complex Number equals Conjugate iff Wholly Real, $\lambda \in \R$.
{{qed}}
Theorem: Existence of Hartogs Number

Let $S$ be a set.
Then $S$ has a Hartogs number.

Proof:
From Hartogs' lemma there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.
We also have that Ordinals are Well-Ordered.
It follows from the definition of well-ordering that there exists a smallest such ordinal.
Hence the result.
{{qed}}
Theorem: Existence of Hartogs Number

Let $S$ be a set.
Then $S$ has a Hartogs number.

Proof:
Follows immediately from Cardinal Equal to Collection of All Dominated Ordinals.
The collection of all dominated ordinals:
:$\set {y \in \On: y \preccurlyeq S}$
is the Hartogs number of $S$.
{{qed}}
Theorem: Left and Right Inverse Mappings Implies Bijection

Let $f: S \to T$ be a mapping.
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are mappings.
Then $f$ is a bijection.

Proof:
Suppose:
:$\exists g_1: T \to S: g_1 \circ f = I_S$
:$\exists g_2: T \to S: f \circ g_2 = I_T$
We have that the Identity Mapping is Bijection, thus $I_S$ and $I_T$ are both bijections.
From Injection if Composite is Injection, if $I_S = g_1 \circ f$ is an injection, then so is $f$.
From Surjection if Composite is Surjection, if $I_T = f \circ g_2$ is a surjection, then so is $f$.
So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.
{{qed}}
Theorem: Left and Right Inverse Mappings Implies Bijection

Let $f: S \to T$ be a mapping.
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are mappings.
Then $f$ is a bijection.

Proof:
Suppose:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
From Injection iff Left Inverse, it follows that $f$ is an injection.
From Surjection iff Right Inverse, it follows that $f$ is a surjection.
So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.
Theorem: Left and Right Inverses of Mapping are Inverse Mapping

Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$.

Proof:
{{begin-eqn}}
{{eqn | l = g_2
      | r = I_S \circ g_2
      | c = {{Defof|Identity Mapping}}
}}
{{eqn | r = \paren {g_1 \circ f} \circ g_2
      | c = by hypothesis
}}
{{eqn | r = g_1 \circ \paren {f \circ g_2}
      | c = Composition of Mappings is Associative
}}
{{eqn | r = g_1 \circ I_T
      | c = by hypothesis
}}
{{eqn | r = g_1
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
From Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection.
It follows from Composite of Bijection with Inverse is Identity Mapping that $g_1 = g_2 = f^{-1}$.
{{qed}}