avewright/proofwiki-math huggingface dataset:
Theorem: Eigenvalues of Triangular Matrix

The eigenvalues of a square triangular matrix are its diagonal elements.

Proof:
Let $\mathbf T$ be a square triangular matrix.
From Characteristic Polynomial of Triangular Matrix, we have:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$ are the diagonal elements of $\mathbf T$.
The eigenvalues are the roots of the characteristic polynomial of $\mathbf T$.
Thus, the eigenvalues of $\mathbf T$ are its diagonal elements.
{{qed}}
Category:Triangular Matrices
Category:Eigenvalues of Square Matrices
ooihg6zmuycre6ev763yfs7wzw5am86
Theorem: Trace of Matrix is Sum of Eigenvalues

Let $\mathbf A$ be a matrix of order $n$ over $\mathbb C$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be the eigenvalues of $\mathbf A$ including algebraic multiplicity.
Then:
:$\ds \map \tr {\mathbf A} = \sum_{i \mathop = 1}^n \lambda_i$
where $\map \tr {\mathbf A}$ is the trace of $\mathbf A$.

Proof:
From Existence of Schur Decomposition for Square Matrix, $\mathbf A$ is similar to an upper triangular matrix $\mathbf T$.
Let $\omega_1, \omega_2, \ldots, \omega_n$ be the eigenvalues of $\mathbf T$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map \tr {\mathbf A}
      | r = \map \tr {\mathbf T}
      | c = Similar Matrices have same Traces
}}
{{eqn | r = \sum_{i \mathop = 1}^n t_{ii}
      | c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \omega_i
      | c = Eigenvalues of Triangular Matrix
}}
{{eqn | r = \sum_{i \mathop = 1}^n \lambda_i
      | c = Similar Matrices have Same Eigenvalues
}}
{{end-eqn}}
{{qed}}
Theorem: Hermitian Conjugate of Normal Matrix is Normal

The Hermitian conjugate of a normal matrix is normal.

Proof:
Let $\mathbf N$ be a normal matrix.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf N \mathbf N^\dagger
      | r = \mathbf N^\dagger \mathbf N
      | c = {{Defof|Normal Matrix}}
}}
{{eqn | l = \paren {\mathbf N^\dagger}^\dagger \paren {\mathbf N^\dagger}
      | r = \paren {\mathbf N^\dagger} \paren {\mathbf N^\dagger}^\dagger
      | c = Hermitian Conjugate is Involution
}}
{{end-eqn}}
Therefore, $\mathbf N^\dagger$ is normal.
{{qed}}
Category:Normal Matrices
Category:Hermitian Conjugates
hh16ah9g6b3fwtfzvyvyrwzyzyld458
Theorem: Lower Triangular Normal Matrix is Diagonal

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

Proof:
Let $\mathbf L$ be a square matrix that is both lower triangular and normal.
From Transpose of Upper Triangular Matrix is Lower Triangular, $\mathbf L^\intercal$ is upper triangular.
Since $\bar 0 = 0$, $\overline {\mathbf L^\intercal} = \mathbf L^\dagger$ is upper triangular.
From Hermitian Conjugate of Normal Matrix is Normal, $\mathbf L^\dagger$ is normal.
From Upper Triangular Normal Matrix is Diagonal, $\mathbf L^\dagger$ is diagonal.
Since $\bar 0 = 0$, $\overline {\mathbf L^\dagger} = \mathbf L^\intercal$ is diagonal.
From Transpose of Transpose of Matrix and Diagonal Matrix is Symmetric, $\mathbf L^\intercal = \mathbf L$.
Thus, $\mathbf L$ is diagonal.
{{qed}}
Category:Triangular Normal Matrix is Diagonal
Category:Normal Matrices
Category:Lower Triangular Matrices
Category:Diagonal Matrices
2j2gopkg9bexkzafninfho2cv5os4a5
Theorem: Triangular Normal Matrix is Diagonal

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

Proof:
=== Upper Triangular Normal Matrix is Diagonal ===
{{:Upper Triangular Normal Matrix is Diagonal}}{{qed|lemma}}
Theorem: (p implies q) implies q, q implies p therefore p/Theorem Form

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

Proof:
{{BeginTableau|\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p}}
{{Assumption|1|\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} }}
{{TheoremIntro|2|\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}|Hypothetical Syllogism }}
{{Substitution|3||\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} \implies \paren {p \implies q} \implies p|2|p|\paren {p \implies q}|r|p}}
{{ModusPonens|4|1|\paren {p \implies q} \implies p|1|3}}
{{TheoremIntro|5|\paren {\paren {p \implies q} \implies p} \implies p|Peirce's Law }}
{{ModusPonens|6|1|p|4|5}}
{{Implication|7||\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p|1|6}}
{{EndTableau|qed}}
Theorem: Finite Infimum in Subframe Equals Infimum in Frame

Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Let $F \subseteq T$ be finite.
The infimum ${\bigwedge}_M F$ of $F$ exists in $M$ and equals the infimum $\bigvee F$ in $L$.

Proof:
{{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From Infimum in Ordered Subset:
:$\forall $ finite $F \subseteq T$, the infimum ${\bigwedge}_M F$ of $F$ exists in $M$ and ${\bigvee}_M F {{=}} \bigvee F$
{{qed}}
Category:Subframes
Category:Frames
fm06rx5ji40p569es64ppu97uj9312n
Theorem: Increasing Sum of Squares of Binomial Coefficients

:$\ds \sum_{r \mathop = 1}^n r \binom n r^2 = \dfrac {\paren {2 n - 1}!} {\paren {\paren {n - 1}!}^2}$
where $\dbinom n r$ denotes the binomial coefficient:
:$\dbinom n r = \dfrac {n!} {r! \paren {n - r} }$

Proof:
We note that $\ds \sum_{r = 1}^n r \binom n r^2$ is the same as $\ds \sum_{r = 0}^n r \binom n r^2$ because the zeroth term is $0 \dbinom n 0^2 = 0$.
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = \sum_{r \mathop = 0}^n r \binom n r^2
      | r = 0 \binom n 0^2 + 1 \binom n 1^2 + 2 \binom n 2^2 + 3 \binom n 3^2 + \cdots + \paren {n - 2} \binom n {n - 2}^2 + \paren {n - 1} \binom n {n - 1}^2 + n \binom n n^2
      | c = 
}}
{{eqn | n = 2
      | l = \sum_{r \mathop = 0}^n r \binom n r^2
      | r = 0 \binom n n^2 + 1 \binom n {n - 1}^2 + 2 \binom n {n - 2}^2 + 3 \binom n {n - 3}^2 + \cdots + \paren {n - 2} \binom n 2^2 + \paren {n - 1} \binom n 1^2 + n \binom n 0^2
      | c = Symmetry Rule for Binomial Coefficients: $\dbinom n r = \dbinom n {n - r}$
}}
{{eqn | ll= \leadsto
      | l = 2 \sum_{r \mathop = 0}^n r \binom n r^2
      | r = n \binom n 0^2 + n \binom n 1^2 + n \binom n 2^2 + n \binom n 3^2 + \cdots + n \binom n {n - 2}^2 + n \binom n {n - 1}^2 + n \binom n n^2
      | c = adding $(1)$ to $(2)$
}}
{{eqn | r = n \paren {\binom n 0^2 + \binom n 1^2 + \binom n 2^2 + \binom n 3^2 + \cdots + \binom n {n - 2}^2 + \binom n {n - 1}^2 + \binom n n^2}
      | c = 
}}
{{eqn | r = n \sum_{r \mathop = 0}^n \binom n r^2
      | c = 
}}
{{eqn | r = n \dbinom {2 n} n
      | c = Sum of Squares of Binomial Coefficients
}}
{{eqn | ll= \leadsto
      | l = \sum_{r \mathop = 1}^n r \binom n r^2
      | r = \dfrac n 2 \dfrac {\paren {2 n}!} {\paren {n!}^2}
      | c = {{Defof|Binomial Coefficient}}: $\dbinom n r = \dfrac {n!} {r! \paren {n - r} }$
}}
{{eqn | r = \dfrac {n^2 \paren {2 n}!} {2 n \paren {n!}^2}
      | c = multiplying top and bottom by $n$
}}
{{eqn | r = \dfrac {\paren {2 n - 1}!} {\paren {\paren {n - 1}!}^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Image of Frame under Frame Homomorphism is Frame

Let $L_1 = \struct {S_1, \preceq_1}$ and $L_2 = \struct {S_2, \preceq_2}$ be frames.
Let $\phi: S_1 \to S_2$ be a frame homomorphism.
Let $L_3 = \struct{S_3, \preceq_3}$ be the ordered set where:
:$S_3$ denotes the image of $S_1$ under $\phi$
:$\preceq_3$ denotes the restricted ordering of $\preceq_2$ to the subset $S_3$.
Then:
:$L_3$ is a subframe of $L_2$.

Proof:
=== {{Lemma|Image of Frame under Frame Homomorphism is Frame|1|Lemma 1}} ===
{{:Image of Frame under Frame Homomorphism is Frame/Lemma 1}}{{qed|lemma}}
Theorem: Image of Frame under Frame Homomorphism is Frame/Lemma 1

Let $T \subseteq S_3$.
Let $\bigvee T$ denote the supremum of $T$ in $L_2$.
Then:
:$\bigvee T \in S_3$.

Proof:
{{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of image:
:$S_3 = \phi \sqbrk {S_1}$
From {{Corollary|Image of Preimage under Mapping}}:
:$(1) \quad T = \phi \sqbrk {\phi^{-1} \sqbrk T}$
Let $S = \phi^{-1} \sqbrk T$.
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {\bigvee S}
      | r = \bigvee {\phi \sqbrk S}
      | c = {{Defof|Frame Homomorphism}}
}}
{{eqn | r = \bigvee {\phi \sqbrk {\phi^{-1} \sqbrk T} }
      | c = Definition of $S$
}}
{{eqn | r = \bigvee T
      | c = By $(1)$
}}
{{end-eqn}}
By definition of image:
:$\bigvee T \in S_3$
{{qed}}
Category:Image of Frame under Frame Homomorphism is Frame
sv85uejw56i33fmnoeorufhga638i69
Theorem: Image of Frame under Frame Homomorphism is Frame/Lemma 2

Let $F \subseteq S_3$ be a finite subset.
Let $\bigwedge F$ denote the infimum of $F$ in $L_2$.
Then:
:$\bigwedge F \in S_3$.

Proof:
{{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of image:
:$\forall y \in F : \exists x \in S_1 : \map \phi x = y$
For each $y \in F$ choose $x_y \in S_1 : \map \phi {x_y} = y$.
Let $G = \set{x_y : y \in F}$.
By definition of image:
:$(1):\quad \phi \sqbrk G = F$
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {\bigwedge G}
      | r = \bigwedge {\phi \sqbrk G}
      | c = {{Defof|Frame Homomorphism}}
}}
{{eqn | r = \bigwedge F
      | c = By $(1)$
}}
{{end-eqn}}
By definition of image:
:$\bigwedge F \in S_3$
{{qed}}
Category:Image of Frame under Frame Homomorphism is Frame
csemtx5y7cue6scpu0va217juzoyq3j
Theorem: Normal Matrix has Orthogonal Eigenspaces

Let $\mathbf N$ be a normal matrix.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf N$.
Then, the eigenspaces of $\lambda_1$ and $\lambda_2$ are orthogonal:
:$\map \ker {\mathbf N - \lambda_1 \mathbf I} \perp \map \ker {\mathbf N - \lambda_2 \mathbf I}$
where:
:$\ker$ denotes kernel
:$\mathbf I$ denotes the unit matrix
:$\perp$ denotes orthogonality.

Proof:
Follows immediately from Normal Matrix is Normal Operator and Eigenvalues of Normal Operator have Orthogonal Eigenspaces
{{qed}}
Category:Normal Matrices
Category:Eigenspaces
fte1v1yw5dz4nzxkd9c2l4vks7vbz81
Theorem: Inclusion Mapping of Subframe into Frame is Frame Homomorphism

Let $L = \struct {S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct {T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
The inclusion mapping $i: M \to L$ is a frame homomorphism.

Proof:
{{Recall|theorem = Supremum in Subframe Equals Supremum in Frame}}
{{:Supremum in Subframe Equals Supremum in Frame}}
{{Recall|theorem = Finite Infimum in Subframe Equals Infimum in Frame}}
{{:Finite Infimum in Subframe Equals Infimum in Frame}}
We have:
{{begin-eqn}}
{{eqn | q = \forall A \subseteq T
      | l = \map i {\bigvee A} 
      | r = \bigvee A
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \bigvee i \sqbrk A
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | q = \forall \text{ finite } F \subseteq T
      | l = \map i {\bigwedge F} 
      | r = \bigwedge F
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \bigwedge i \sqbrk F
      | c = {{Defof|Inclusion Mapping}}
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Subframes
Category:Frame Homomorphisms
Category:Inclusion Mappings
q92hfpz271xv6ca28031xio0qgddycw
Theorem: Supremum in Subframe Equals Supremum in Frame

Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Let $A \subseteq T$.
The supremum ${\bigvee}_M A$ of $A$ exists in $M$ and equals the supremum $\bigvee A$ in $L$.

Proof:
{{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From Supremum in Ordered Subset:
:$\forall A \subseteq T$, the supremum ${\bigvee}_M A$ of $A$ exists in $M$ and ${\bigvee}_M A {{=}} \bigvee A$
{{qed}}
Category:Subframes
Category:Frames
p3ghvw0nb6ahqffgw770zo3k2htzyyh
Theorem: Subframe is a Frame

Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Then $M$ is a frame.

Proof:
{{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
Theorem: Simple Spectrum Approximation of Normal 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 U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf A
}}
{{end-eqn}}
{{qed}}
Category:Simple Spectrum Approximation of Normal Matrix
9ikjtwsrk54q2ommh9xft6xaa5e41fr
Theorem: Simple Spectrum Approximation of Normal Matrix/Property 2

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

Proof:
Let $\mathbf D = \mathbf U^\dagger \mathbf D \mathbf U$ be the diagonalized version of $\mathbf A$.
We have:
{{begin-eqn}}
{{eqn | l = \map {\mathbf A} t
      | r = \mathbf A + t \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf U \mathbf D \mathbf U^\dagger + t \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf U \paren {\mathbf D + t \mathbf E} \mathbf U^\dagger
}}
{{end-eqn}}
Since $\mathbf D$ and $t \mathbf E$ are diagonal, $\mathbf D + t \mathbf E$ is diagonal.
From Spectral Theorem for Normal Matrices, since $\map {\mathbf A} t$ is unitarily similar to a diagonal matrix over $\mathbb C$, $\map {\mathbf A} t$ is normal for all values of $t$.
{{qed}}
Category:Simple Spectrum Approximation of Normal Matrix
eepjzc5ci8flnbwrohmz0o39f9pkzy1
Theorem: Simple Spectrum Approximation of Normal Matrix/Property 3

:$\map {\mathbf A} t$ has the same eigenvectors as $\mathbf A$ for all values of $t$

Proof:
From Property $(2)$ of Simple Spectrum Approximation of Normal Matrix, $\map {\mathbf A} t$ is normal.
From Spectral Theorem for Normal Matrices, $\map {\mathbf A} t$ is diagonalizable.
From Simple Spectrum Approximation of Diagonalizable Matrix, $\map {\mathbf A} t$ has the same eigenvectors as $\mathbf A$.
{{qed}}
Category:Simple Spectrum Approximation of Normal Matrix
rl9bf2q2x0jqewj0l9y1leeoqp5i1nc
Theorem: Simple Spectrum Approximation of Normal Matrix/Property 5

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

Proof:
From Property $(2)$ of Simple Spectrum Approximation of Normal Matrix, $\map {\mathbf A} t$ is normal.
From Spectral Theorem for Normal Matrices, $\map {\mathbf A} t$ is diagonalizable.
From Simple Spectrum Approximation of Diagonalizable Matrix, there exists $\tau > 0$ such that:
:$0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is simple.
{{qed}}
Category:Simple Spectrum Approximation of Normal Matrix
1lggl63gs9r1liutb1kn0ul4lb2xxfb
Theorem: Halting Problem is Undecidable

No Turing machine can be programmed to determine whether an arbitrary Turing machine will halt on arbitrary input.

Proof:
{{AimForCont}} there exists a Turing machine $\map H {T, I}$ that determines whether another Turing machine $T$ halts with input $I$:
:$\map H {T, I} = \begin{cases} \text{true} & : T \text{ halts on input } I \\ \text{false} & : T \text{ never halts on input } I \end{cases}$
Then, the following Turing machine can be implemented:
:$\map R T = \begin{cases} \text{halt} & : \map H {T, T} \text{ is false} \\ \text{run forever} & : \map H {T, T} \text{ is true} \end{cases}$
That is, $\map R T$ halts immediately when $T$ fails to halt when run with its own machine description as input.
Otherwise, $\map R T$ never halts.
Now, consider what will happen if $R$ is run on itself:
:$\map R R = \begin{cases} \text{halt} & : \map H {R, R} \text{ is false} \\ \text{run forever} & : \map H {R, R} \text{ is true} \end{cases}$
However, if $\map H {R, R}$ is false, then, according to the definition of $H$, $\map R R$ never halts, which contradicts the instruction to $\text{halt}$ in the definition of $R$.
Furthermore, if $\map H {R, R}$ is true, then, according to the definition of $H$, $\map R R$ halts, which also contradicts the instruction to $\text{run forever}$ in the definition of $R$.
These contradictions mean that our assumption that $\map H {T, I}$ exists must be false.
{{qed}}
Category:Halting Problem
Category:Turing Machines
3iddty5t7r2yl55y4a79k4ubsdqv55z
Theorem: Sum from 0 to Infinity of (r+1) x^r

Let $x \in \R$ such that $\size x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 0}^\infty \paren {r + 1} x^r
      | r = 1 + 2 x + 3 x^2 + 4 x^3 + \cdots
      | c = 
}}
{{eqn | r = \dfrac 1 {\paren {1 - x}^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Sum from $0$ to $n - 1$ of $\paren {r + 1} x^r$:
:$\ds \sum_{r \mathop = 0}^{n - 1} \paren {r + 1} x^r = \dfrac {1 - x^n} {\paren {1 - x}^2} - \dfrac {n x^n} {1 - x}$
From Real Null Sequence: $n^\alpha x^n$:
:$\ds \lim_{n \mathop \to \infty} n x^n = 0$
Then we have that:
:$\ds \lim_{n \mathop \to \infty} x^n = 0$
{{explain|I can't find the link for the above result, basic as it is}}
The result follows.
{{qed}}
Theorem: Dirac Bra Grouping is Conjugate Linear

Let $\bra \alpha$ and $\bra \beta$ be bras in some dual space.
Let $h$ and $k$ be scalars.
Then:
:$\bra {h \alpha + k \beta} = \bra \alpha h^\dagger + \bra \beta k^\dagger$
where $h^\dagger$ is the complex conjugate of $h$.

Proof:
{{begin-eqn}}
{{eqn | l = \bra {h \alpha + k \beta}
      | r = \ket {h \alpha + k \beta}^*
      | c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {h \ket \alpha + k \ket \beta}^*
      | c = {{Defof|Dirac Notation/Ket/Abuse of Notation|Ket notation}}
}}
{{eqn | r = h^\dagger \ket \alpha^* + k^\dagger \ket \beta^*
      | c = Adjoint is Conjugate Linear
}}
{{eqn | r = \bra \alpha h^\dagger + \bra \beta k^\dagger
      | c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{end-eqn}}
{{qed}}
Category:Bras
Category:Dirac Notation
qv64hi3wzfn1e244ffyl9legghau6x7
Theorem: Linear Transform on Dirac Bra

Let $\bra x$ be a bra in some dual space.
Let $T$ be a linear transform.
Then:
:$\bra x T = \bra {T^* x}$

Proof:
{{begin-eqn}}
{{eqn | l = \bra x T
      | r = \ket x^* T
      | c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {T^* \ket x}^*
      | c = Adjoint of Composition of Linear Transformations is Composition of Adjoints
}}
{{eqn | r = \ket {T^* x}^*
      | c = {{Defof|Dirac Notation/Ket/Abuse of Notation|Ket notation}}
}}
{{eqn | r = \bra {T^* x}
      | c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{end-eqn}}
{{qed}}
Category:Bras
Category:Dirac Notation
7m2c5raqes3zxj3bxj4f9sxj8mmaqsy
Theorem: Dirac Inner Product is Conjugate Linear in First Argument

Let $\ket \alpha$, $\ket \beta$, and $\ket \gamma$ be kets.
Let $k$ be a scalar.
Then:
:$\braket {k \alpha + \beta} \gamma = \bar k \braket \alpha \gamma + \braket \beta \gamma$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \braket {k \alpha + \beta} \gamma
      | r = \overline {\braket \gamma {k \alpha + \beta} }
      | c = Inner product is conjugate symmetric
}}
{{eqn | r = \overline {k \braket \gamma \alpha + \braket \gamma \beta}
      | c = Dirac inner product is linear in the second argument
}}
{{eqn | r = \bar k \overline {\braket \gamma \alpha} + \overline {\braket \gamma \beta}
      | c = Sum of Complex Conjugates and Product of Complex Conjugates
}}
{{eqn | r = \bar k \braket \alpha \gamma + \braket \beta \gamma
      | c = Inner product is conjugate symmetric
}}
{{end-eqn}}
{{qed}}
Category:Dirac Notation
Category:Inner Products
Category:Quantum Mechanics
s0a1m1jso0d4fuwrwfbub7gt466xl0o
Theorem: Power Series Expansion for Logarithm of m over n

{{begin-eqn}}
{{eqn | l = \map \ln {\frac m n}
      | r = 2 \sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \paren {\frac {m - n} {m + n} }^{2 n + 1}
}}
{{eqn | r = 2 \paren {\frac {m - n} {m + n} + \frac 1 3 \paren {\frac {m - n} {m + n} }^3 + \frac 1 5 \paren {\frac {m - n} {m + n} }^5 + \frac 1 7 \paren {\frac {m - n} {m + n} }^7 + \cdots}
}}
{{end-eqn}}
valid for all $m, n \in \R_{>0}$.

Proof:
From Power Series Expansion for $\map \ln {\dfrac {1 + x} {1 - x} }$:
:$(1): \quad \ds \map \ln {\dfrac {1 + x} {1 - x} } = 2 \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots}$
for $-1 < x \le 1$.
Let $m, n \in \R_{>0}$.
Let $x = \dfrac {m - n} {m + n}$.
Then as $m$ and $n$ are both positive:
:$0 < x < 1$
and we have:
:$\dfrac m n = \dfrac {1 + x} {1 - x}$
The result follows by substituting $\dfrac {m - n} {m + n}$ for $x$ in $(1)$.
{{qed}}
Theorem: Sum from 1 to Infinity of x^r over r(r+1)

{{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }
      | r = \dfrac x {1 \times 2} + \dfrac {x^2} {2 \times 3} + \dfrac {x^3} {3 \times 4} + \cdots
      | c = 
}}
{{eqn | r = \paren {\dfrac 1 x - 1} \map \ln  {1 - x} + 1
      | c = 
}}
{{end-eqn}}
valid for $-1 \le x < 1$.

Proof:
Let:
:$S = \ds \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }$
From Partial Fractions Expansion: $\dfrac 1 {x \paren {x + 1} }$:
:$\dfrac 1 {r \paren {r + 1} } = \dfrac 1 r - \dfrac 1 {r + 1}$
and so:
:$\dfrac {x^r} {r \paren {r + 1} } = \dfrac {x^r} r - \dfrac {x^r} {r + 1}$
Hence:
{{begin-eqn}}
{{eqn | l = S
      | r = \sum_{r \mathop = 1}^\infty \paren {\dfrac {x^r} r - \dfrac {x^r} {r + 1} }
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \dfrac {x^r} r - \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r + 1}
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \dfrac {x^r} r - \dfrac 1 x \sum_{r \mathop = 1}^\infty \dfrac {x^{r + 1} } {r + 1}
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \dfrac {x^r} r - \dfrac 1 x \sum_{r \mathop = 2}^\infty \dfrac {x^r} r
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \sum_{r \mathop = 1}^\infty \dfrac {x^r} r - \dfrac 1 x \paren {\sum_{r \mathop = 1}^\infty \dfrac {x^r} r - x}
      | c = 
}}
{{eqn | r = -\map \ln {1 - x} - \dfrac 1 x \paren {-\map \ln {1 - x} - x}
      | c = Power Series Expansion for $\map \ln {1 - x}$ (twice)
      | cc= provided $-1 \le x < 1$
}}
{{eqn | r = \paren {\dfrac 1 x - 1} \map \ln  {1 - x} + 1
      | c = simplifying
      | cc= 
}}
{{end-eqn}}
{{qed}}
Theorem: Polarization Identity/Complex Vector Space/Corollary

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\C$. 
Let $\norm \cdot$ be the inner product norm on $V$. 
Then, we have: 
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$
for each $x, y \in V$.

Proof:
From Inner Product is Sesquilinear, the mapping $q : V \times V \to \C$ defined by:
:$\map q {x, y} = \innerprod x y$ for each $\tuple {x, y} \in V \times V$
is sesquilinear.
Define $Q : V \to \R$ by:
:$\map Q x = \innerprod x x$ for each $x \in V$.
Let $\norm {\, \cdot \,}$ be the inner product norm of $\innerprod \cdot \cdot$.
Then $\map Q x = \norm x^2$ for each $x \in V$. 
Hence, applying Polarization Identity: Complex Vector Space with $q = \innerprod \cdot \cdot$, we obtain:
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$ for each $x, y \in V$. 
{{qed}}
Category:Polarization Identity
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