nyuuzyou/wikis huggingface dataset:
Hensel's Lemma/P-adic Integers/Lemma 1

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\map {F'} X$ be the (formal) derivative of $F$.
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
:$\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
:$\map {F'} {\alpha_0} \not\equiv 0 \pmod {p\Z_p}$
where $p\Z_p$ denotes the principal ideal of $\Z_p$ generated by $p$.
Then:
:There exists a unique $p$-adic expansion $\ds \sum_{n = 0}^\infty d_n p^n$:
::$\forall k : a_k = \ds \sum_{n = 0}^k d_n p^n$ satisfies:
:::$(1) \quad \map F {a_k} \equiv 0 \pmod {p^{k+1}\Z_p}$
:::$(2) \quad a_k \equiv \alpha_0 \pmod {p\Z_p}$
where $p^k\Z_p$ denotes the principal ideal of $\Z_p$ generated by $p^k$.
== Proof ==
The Second Principle of Recursive Definition is used to construct the sequence
$\sequence {d_n}$.
Let $T$ be the set of $p$-adic digits.
For each $k \in \N_{>0}$, let:
:$S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\ds \sum_{n = 0}^{k-1} b_n p^n } \equiv 0 \pmod{p^k\Z_p} \quad \text{and} \quad \ds \sum_{n = 0}^{k-1} b_n p^n \equiv \alpha_0 \pmod{p\Z_p}}$
Let $d_0$ be the first $p$-adic digit of the canonical expansion of $\alpha_0$.
=== Lemma 4 ===
=== Lemma 5 ===
For $k \in \N_{>0}$, let $G_k : T^k \to T$ be the mapping defined by:
:$\map {G_k} {b_0, b_1, \ldots, b_{k-1}} = \begin{cases}
\map b {b_0, b_1, \ldots, b_{k-1}} & : \tuple {b_0, b_1, \ldots, b_{k-1}} \in S_k\\
0 & : \tuple {b_0, b_1, \ldots, b_{k-1}} \notin S_k
\end{cases}$
From Lemma 5:
:$G_k$ is well-defined for all $k \in \N$
and
:$\forall \tuple {b_0, b_1, \ldots, b_{k-1}} \in S_k$:
::$\tuple {b_0, b_1, \ldots, b_{k-1}, \map {G_k} {b_0, b_1, \ldots, b_k} } \in S_{k+1}$
From Second Principle of Recursive Definition, there exists exactly one mapping $d: \N \to T$ such that:
:$\forall x \in \N: \map d x = \begin{cases}
d_0 & : x = 0 \\
\map {G_n} {\map d 0, \ldots, \map d n} & : x = n + 1
\end{cases}$
We have:
:$(a)\quad \tuple{d_0} \in S_0$
:$(b)\quad \tuple{d_0, d_1, \ldots, d_{k-1}} \in S_k \implies \tuple{d_0, d_1, \ldots, d_k} \in S_{k+1}$
From Principle of Mathematical Induction:
:$\forall k \in N_{>0}: \tuple{d_0, d_1, \ldots, d_{k-1}} \in S_k$
That is, for all $k \in N$:
:$a_k = \ds \sum_{n = 0}^k d_n p_n$ satisfies:
::$(a) \quad \map F {a_k} \equiv 0 \pmod {p^{k+1}\Z_p}$
::$(b) \quad a_k \equiv \alpha_0 \pmod {p\Z_p}$
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 10

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then:
:$\forall x \in \Z_p: p^k x \equiv 0 \pmod{p^{k+1}\Z_p} \implies x \equiv 0 \pmod{p\Z_p}$
== Proof ==
We have:
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 2

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\alpha \in \Z_p$ be a $p$-adic number with $p$-adic expansion:
:$\alpha = \ds \sum_{n \mathop = 0}^\infty d_n p^n$
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer.
Let $p \Z_p$ denote the principal ideal of $\Z_p$ generated by $p$.
For all $x, y \in \Z_p$, let:
:$x \equiv y \pmod {p \Z_p}$
denote congruence modulo the ideal $p \Z_p$.
For all $k \in \N$, let the partial sum $a_k = \ds \sum_{n \mathop = 0}^k d_n p_n$ satisfy:
:$a_k \equiv \alpha_0 \pmod {p \Z_p}$
Then:
:$\alpha \equiv \alpha_0 \pmod {p\Z_p}$
== Proof ==
From Ideals of P-adic Integers:
:$p \Z_p$ is an ideal of the ring $\Z_p$
By :
:$\forall k \in \N: a_k - \alpha_0 \in p \Z_p$
By :
:$\alpha = \ds \lim_{k \mathop \to \infty} a_k$
From Sum Rule for Sequences in Normed Division Ring:
:$\alpha - \alpha_0 = \ds \lim_{k \mathop \to \infty} a_k - \alpha_0$
From Closed Subgroups of P-adic Integers:
:$p \Z_p$ is a closed set in the $p$-adic metric
From Subset of Metric Space contains Limits of Sequences iff Closed:
:$\alpha - \alpha_0 \in p \Z_p$
By :
:$\alpha \equiv \alpha_0 \pmod {p \Z_p}$
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 4

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
:$\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
Let $T$ be the set of $p$-adic digits.
Let:
:$S_1 = \set{\tuple{b_0} \subseteq T^1 : \map F {b_0} \equiv 0 \pmod{p\Z_p} \land b_0 \equiv \alpha_0 \pmod{p\Z_p}}$
Let $d_0$ be the first $p$-adic digit of the canonical expansion of $\alpha_0$.
Then:
:$\tuple{d_0} \in S_1$
== Proof ==
Let the $p$-adic expansion for $\alpha_0$ be:
:$\alpha_0 = \ds \sum_{n = 0}^\infty d_n p^n$
We have:
From Polynomials of Congruent Ring Elements are Congruent:
:$\map F {d_0} \equiv \map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
Hence:
:$\tuple {d_0} \in S_1$
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 5

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\map {F'} X$ be the (formal) derivative of $F$.
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
:$\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
:$\map {F'} {\alpha_0} \not\equiv 0 \pmod {p\Z_p}$
Let $T$ be the set of $p$-adic digits.
For each $k \in \N_{>0}$, let:
:$S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\ds \sum_{n = 0}^{k-1} b_n p^n} \equiv 0 \pmod{p^k\Z_p} \quad \text{and} \quad \ds \sum_{n = 0}^{k-1} b_n p^n \equiv \alpha_0 \pmod{p\Z_p}}$
Let:
:$\tuple{b_0, b_1, \ldots, b_{k-1}} \in S_k$.
Then there exists a unique $p$-adic digit $\map b {b_0, b_1, \ldots, b_{k-1}}$:
:$\tuple{b_0, b_1, \ldots, b_{k-1}, \map b {b_0, b_1, \ldots, b_{k-1}}} \in S_{k+1}$
== Proof ==
=== Lemma 6 ===
=== Lemma 7 ===
=== Lemma 8 ===
=== Lemma 9 ===
Let:
:$a = \ds \sum_{n = 0}^{k-1} b_n p_n$
From Lemma 6:
:$\exists \beta \in \Z_p : \map F a = \beta p^k$
By hypothesis:
:$a \equiv \alpha_0 \pmod{p\Z_p}$
From Polynomials of Congruent Ring Elements are Congruent:
:$\map {F'} a \equiv \map {F'} {\alpha_0} \pmod{p\Z_p}$
By hypothesis:
:$\map {F'} {\alpha_0} \not\equiv 0 \pmod{p\Z_p}$
Hence:
:$\map {F'} a \not\equiv 0 \pmod{p\Z_p}$
It follows that:
:$\map {F'} a \ne 0$
From Lemma 7:
:$\exists b_k \in T : b_k p^k \equiv \dfrac {-\beta p^k} {\map {F'} a} \pmod{p^{k+1}\Z_p}$
Let $a' = a + b_kp^k = \ds \sum_{n = 0}^k b_n p^n$.
We have:
We have:
 
It has been shown:
:$\tuple{b_0, b_1, \ldots, b_{k-1}, b_k} \in S_{k+1}$
From Lemma 9:
:$b_k$ is unique.
Let $\map b {b_0, b_1, \ldots, b_{k-1}} = b_k$.
The result follows.
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 6

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $x \in \Z_p$.
Let $k \in \N_{>0}$.
Then:
:$x \equiv 0 \pmod {p^k\Z_p} \implies \exists y \in \Z_p : x = y p^k$
== Proof ==
Let:
:$x \equiv 0 \pmod{p^k\Z_p}$
By definition of congruence modulo an ideal:
:$x \in p^k\Z_p$
By definition of principal ideal:
:$\exists y \in \Z_p : x = y p^k$
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 7

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $T$ be the set of $p$-adic digits.
Let $k \in \N_{>0}$.
Then:
:$x \in \Z_p \implies \exists y \in T : y p^k \equiv x p^k \pmod {p^{k+1}\Z_p}$
== Proof ==
Let $x \in \Z_p$.
From P-adic Integer is Limit of Unique P-adic Expansion, let:
:$x = \ds \sum_{n \mathop = 0}^\infty d_n p^n$
We have:
Let $y = d_0$.
The result follows.
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 8

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\map {F'} X$ be the (formal) derivative of $F$.
Let $k \in \N_{>0}$.
Then:
:$x, y \in \Z_p \implies \map F {x + y p ^k} \equiv \map F x + y p^k \map {F'} x \pmod {p^{k+1}\Z_p}$
== Proof ==
Let $\map F X = \ds \sum_{j \mathop = 0}^r c_j X^j$ where $X$ is the indeterminate and $c_0, c_1, \ldots, c_r \in \Z_p$.
Then:
:$\map {F'} X = \ds \sum_{j \mathop = 1}^r j c_j X^j$
We have:
Category:Hensel's Lemma
Hensel's Lemma/P-adic Integers/Lemma 9

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\map {F'} X$ be the (formal) derivative of $F$.
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
:$\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
:$\map {F'} {\alpha_0} \not\equiv 0 \pmod {p\Z_p}$
Let $T$ be the set of $p$-adic digits.
For each $k \in \N_{>0}$, let:
:$S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\ds \sum_{n = 0}^{k-1} b_n p^n} \equiv 0 \pmod{p^k\Z_p} \quad \text{and} \quad \ds \sum_{n = 0}^{k-1} b_n p^n \equiv \alpha_0 \pmod{p\Z_p}}$
Let:
:$\tuple{b_0, b_1, \ldots, b_{k-1}} \in S_k$.
Then:
:$\paren{c, d \in T : \tuple{b_0, b_1, \ldots, b_{k-1}, c}, \tuple{b_0, b_1, \ldots, b_{k-1}, d} \in S_{k+1}} \implies c = d$
== Proof ==
=== Lemma 7 ===
=== Lemma 8 ===
=== Lemma 10 ===
Let:
:$c, d \in T : \tuple{b_0, b_1, \ldots, b_{k-1}, c}, \tuple{b_0, b_1, \ldots, b_{k-1}, d} \in S_{k+1}$
Let:
:$a = \ds \sum_{n = 0}^{k-1} b_n p^n$
We have:
From Lemma 7:
:$\exists b \in T : \map {F'} a \equiv b \pmod{p\Z_p}$
We have:
We have:
The result follows.
Category:Hensel's Lemma
Hensel's Lemma for Composite Numbers

== Theorem ==
Let $b \in \Z \setminus \set {-1, 0, 1}$ be an integer.
Let $k > 0$ be a positive integer.
Let $\map f X \in \Z \sqbrk X$ be a polynomial.
Let $x_k \in \Z$ such that:
:$\map f {x_k} \equiv 0 \pmod {b^k}$
:$\gcd \set {\map {f'} {x_k}, b} = 1$
Then for every integer $l \ge 0$, there exists an integer $x_{k + l}$ such that:
:$\map f {x_{k + l} } \equiv 0 \pmod {b^{k + l} }$
:$x_{k + l} \equiv x_k \pmod {b^k}$
and any two integers satisfying these congruences are congruent modulo $b^{k + l}$.
Moreover, for all $l \ge 0$ and any solutions $x_{k + l}$ and $x_{k + l + 1}$:
:$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$
:$x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$
== Proof ==
We use induction on $l$.
The base case $l = 0$ is trivial.
Let $l \ge 0$ be such that a solution $x_{k + l}$ exists and is unique up to a multiple of $b^{k + l}$.
Choose a solution $x_{k + l}$.
Each solution $x_{k + l + 1}$ is also a solution of the previous congruence.
By uniqueness, it has to satisfy $x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$, hence is of the form $x_{k + l} + t b^{k + l}$ with $t \in \Z$.
Let $d = \deg f$.
We have:
Because $\map {f'} {x_{k + l} } \equiv \map {f'} {x_k} \pmod b$, $\map {f'} {x_{k + l} }$ is invertible modulo $b$.
Thus $x_{k + l} + t b^{k + l}$ is a solution modulo $b^{k + l + 1}$ :
:$t \equiv - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod b$
Thus, necessarily:
:$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$
which proves the existence and uniqueness.
By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations:
:$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$
:$x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$
== Also see ==
* Hensel's Lemma
Category:Hensel's Lemma
Heptagonal Number/Sequence

== Definition ==
The sequence of heptagonal numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, \ldots$
== Sources ==
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
* : $81$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
Category:Heptagonal Numbers
Hermite's Formula for Hurwitz Zeta Function

== Theorem == 
:$\ds \map \zeta {s, q} = \frac 1 {2 q^s} + \frac { q^{1 - s} } {s - 1} + 2 \int_0^\infty \frac {\map \sin {s \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 s} \paren {e^{2 \pi x} - 1} } \rd x$
where: 
:$\zeta$ is the Hurwitz zeta function
:$\map \Re s > 1$
:$\map \Re q > 0$.
== Proof ==
To prove this theorem, we can make use of Binet's Second Formula for Log Gamma:
Let $q$ be a complex number with a positive real part.
Then: 
:$\ds \Ln \map \Gamma q = \paren {q - \frac 1 2} \Ln q - q + \frac 1 2 \ln 2 \pi + 2 \int_0^\infty \frac {\map \arctan {x / q} } {e^{2 \pi x} - 1} \rd x$
Applying the $n$th fractional derivative with respect to $q$ on both sides, we get:
Now we can solve the fractional derivatives:
So:
Dividing both sides by the common factor, we get the initial formula:
:$\ds \map \bszeta {n, q} = \frac {q^{1 - n} } {n - 1} + \frac 1 {2 q^n} + 2 \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x$
== Sources ==
* : $13.2$: Hermite's formula for $\map \zeta {s, a}$
Category:Hurwitz Zeta Function
Hermite-Lindemann-Weierstrass Theorem

== Theorem ==
Let $a_1, \cdots, a_n$ be algebraic numbers (possibly complex) that are linearly independent over the rational numbers $\Q$.
Then:
:$e^{a_1}, \cdots, e^{a_n}$ are algebraically independent
where $e$ is Euler's number.
=== Weaker ===
== Proof ==
== Also known as ==
Category:Hermite-Lindemann-Weierstrass Theorem
Category:Transcendental Number Theory
Hermite-Lindemann-Weierstrass Theorem/Also known as

== Hermite-Lindemann-Weierstrass Theorem: Also known as == 
The Hermite-Lindemann-Weierstrass Theorem is also known as:
* the Lindemann-Weierstrass Theorem
* the Hermite-Lindemann Theorem.
Category:Hermite-Lindemann-Weierstrass Theorem
Hermite-Lindemann-Weierstrass Theorem/Weaker

== Theorem ==
Let $a$ be a non-zero algebraic number (possibly complex).
Then:
:$e^a$ is transcendental
where $e$ is Euler's number.
=== Corollary ===
== Proof ==
This follows trivially from Hermite-Lindemann-Weierstrass Theorem by taking $n = 1$.
Category:Hermite-Lindemann-Weierstrass Theorem
Hermite-Lindemann-Weierstrass Theorem/Weaker/Corollary

== Theorem ==
Let $a$ be a algebraic number (possibly complex) which is neither $0$ nor $1$.
Then:
:any value of $\ln a$ is transcendental
where $\ln$ denotes complex natural logarithm.
== Proof ==
 $\ln a$ is not transcendental.
Hence, by definition, it is algebraic.
Since $a$ is not $1$, $\ln a$ cannot be $0$.
Hence, by the weaker Hermite-Lindemann-Weierstrass theorem, $e^{\ln a} = a$ is transcendental.
This contradicts with the assumption that $a$ is algebraic.
Hence, $\ln a$ must be transcendental.
Category:Hermite-Lindemann-Weierstrass Theorem
Hermite Constant for Dimension 2

== Theorem ==
The Hermite constant for dimension $2$ is:
:$\gamma_2 = \dfrac 2 {\sqrt 3}$
or, as it is often presented:
:$\paren {\gamma_2}^2 = \dfrac 4 3$
== Proof ==
The statement of the result to be proved can be expressed as:
:There exist non-zero $x$ and $y$ such that:
::$\paren {a x^2 + 2 b x y + c y^2}^2 \le \size {a c - b^2} \times \dfrac 4 3$
== Sources ==
* : $1,33333 33333 33 \ldots$
Category:Hermite Constants
Hermitian Element of C*-Algebra Decomposes into Positive Elements/Corollary

== Corollary ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $b \in A$.
Then there exists positive elements $b_1, b_2, b_3, b_4 \in A$ such that:
:$b = b_1 - b_2 + i \paren {b_3 - b_4}$
== Proof ==
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists Hermitian elements $\alpha, \beta \in A$ such that:
:$b = \alpha + i \beta$
From Hermitian Element of C*-Algebra Decomposes into Positive Elements, there exists positive elements $b_1, b_2, b_3, b_4 \in A$ such that:
:$\alpha = b_1 - b_2$
and:
:$\beta = b_3 - b_4$
Hence:
:$b = b_1 - b_2 + i \paren {b_3 - b_4}$
Category:Hermitian Element of C*-Algebra Decomposes into Positive Elements
Hermitian Element of Unital C*-Algebra is Linear Combination of Two Unitary Elements

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be Hermitian. 
Then there exists unitary elements $u, v \in A$ and $\alpha \ge 0$ such that:
:$x = \alpha \paren {u + v}$ 
== Proof ==
If $x = {\mathbf 0}_A$, then we can take $\alpha = 0$ and $u = v = {\mathbf 1}_A$.
Take $x \ne {\mathbf 0}_A$.
First take $\norm x \le 1$. 
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have:
:$\norm x = \sup \set {\cmod \lambda : \lambda \in \map {\sigma_A} x} \le 1$
where $\map {\sigma_A} x$ denotes the spectrum of $x$ in $A$.
From Spectrum of Hermitian Element in Unital C*-Algebra is Real we have $\map {\sigma_A} x \subseteq \R$.
We therefore have:
:$\map {\sigma_A} x \subseteq \closedint {-1} 1$
Hence $1 - t^2 \ge 0$ for each $t \in \map {\sigma_A} x$. 
Hence by the Spectral Mapping Theorem for Polynomials, we obtain:
:$\map {\sigma_A} { {\mathbf 1}_A - x^2} = \set {1 - t^2 : t \in \map {\sigma_A} x} \subseteq \hointr 0 \infty$
From Star of Product of Elements in *-Algebra and Identity Element in Unital *-Algebra is Hermitian, we have that ${\mathbf 1}_A - x^2$ is Hermitian.
So ${\mathbf 1}_A - x^2$ is positive. 
From Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra, there exists a positive $s \in A$ with $s^2 = {\mathbf 1}_A - x^2$.
Take $u = x + i s$ and $v = x - i s$.
We have:
We also have:
:$u^\ast u = \paren {x - is} \paren {x + i s} = x^2 + s^2 = {\mathbf 1}_A$
Hence $u$ is unitary.
Since $v = u^\ast$, we obtain that $v$ is unitary also. 
We clearly have:
Hence we have the result for $\norm x \le 1$.
Now take $\norm x > 1$.
We have that:
:$\ds \norm {\frac x {\norm x} } \le 1$
from .
Hence there exists unitary elements $u, v \in A$ such that:
:$\ds \frac x {\norm x} = \frac 1 2 \paren {u + v}$
so that:
:$\ds x = \frac {\norm x} 2 \paren {u + v}$
Hence we are done with $\alpha = \dfrac {\norm x} 2$. 
== Sources ==
* : $2.2$: Positive Elements of $C^\ast$-Algebras
Category:Hermitian Elements of *-Algebras
Category:Unitary Elements of Unital *-Algebras
Hermitian Elements of *-Algebra form Real Vector Subspace

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $A_{\mathbf{SA} }$ be the set of Hermitian elements of $A$.
Then $A_{\mathbf{SA} }$ is a $\R$-vector subspace of $A$.
== Proof ==
We use the One-Step Vector Subspace Test.
From Zero Vector in *-Algebra is Hermitian, we have ${\mathbf 0}_A \in A_{\mathbf{SA} }$.
In particular $A_{\mathbf{SA} } \ne \O$.
Let $a, b \in A_{\mathbf{SA} }$ and $\lambda \in \R$.
We have, from $(\text C^\ast 2)$ and $(\text C^\ast 4)$ in the definition of an involution:
:$\paren {a + \lambda b}^\ast = a^\ast + \lambda b^\ast = a + \lambda b$
Hence $a + \lambda b$ is Hermitian and $a + \lambda b \in A_{\mathbf{SA} }$.
So by the One-Step Vector Subspace Test, we have that $A_{\mathbf{SA} }$ is a $\R$-vector subspace of $A$.
Category:*-Algebras
Hermitian Matrix has Real Eigenvalues/Proof 1

== Theorem ==
== Proof ==
Let $\mathbf A$ be a Hermitian matrix.
Then, by definition:
:$\mathbf A = \mathbf A^\dagger$
where $\mathbf A^\dagger$ denotes the Hermitian conjugate of $\mathbf A$.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$.
By definition of eigenvector:
:$\mathbf{A v} = \lambda \mathbf v$
Left-multiplying both sides by $\mathbf v^*$, we obtain:
:$(1): \quad \mathbf v^* \mathbf {A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$
Firstly, note that both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are $1 \times 1$-matrices.
Now observe that, using Conjugate Transpose of Matrix Product: General Case:
:$\paren {\mathbf v^* \mathbf{A v} }^\dagger = \mathbf v^* \mathbf A^* \paren {\mathbf v^*}^*$
As $\mathbf A$ is Hermitian, and $\paren {\mathbf v^*}^* = \mathbf v$ by Conjugate Transpose is Involution, it follows that:
:$\mathbf v^* \mathbf A^\dagger \paren {\mathbf v^*}^* = \mathbf v^* \mathbf{A v}$
That is, $\mathbf v^* \mathbf {A v}$ is also Hermitian.
By Product with Conjugate Transpose Matrix is Hermitian, $\mathbf v^* \mathbf v$ is Hermitian. 
So both $\mathbf v^* \mathbf {A v}$ and $\mathbf v^* \mathbf v$ are Hermitian $1 \times 1$ matrices. 
Now suppose that we have for some $a,b \in \C$:
:$\mathbf v^* \mathbf {A v} = \begin {bmatrix} a \end {bmatrix}$
:$\mathbf v^* \mathbf v = \begin {bmatrix} b \end {bmatrix}$
Note that $b \ne 0$ as an eigenvector is by definition non-zero.
By definition of Hermitian matrix:
:$\begin {bmatrix} a \end {bmatrix} = \begin {bmatrix} a \end {bmatrix}^*$ and $\begin {bmatrix} b \end {bmatrix} = \begin {bmatrix} b \end {bmatrix}^*$
By definition of Hermitian conjugate:
:$\begin {bmatrix} a \end {bmatrix}^* = \begin {bmatrix} \bar a \end {bmatrix}$ and $\begin {bmatrix} b \end {bmatrix}^* = \begin {bmatrix} \bar b \end {bmatrix}$
where $\bar a$ denotes the complex conjugate of $a$.
So by definition of equality of matrices:
:$a = \bar a$ and $b = \bar b$
By Complex Number equals Conjugate iff Wholly Real:
:$a, b \in \R$, that is, are real.
From equation $(1)$, it follows that:
:$\begin {bmatrix} a \end{bmatrix} = \lambda \begin{bmatrix} b \end{bmatrix}$.
Thus:
:$a = \lambda b$
Hence because $b \ne 0$:
:$\lambda = \dfrac a b$
Hence $\lambda$, being a quotient of real numbers, is real.
Category:Hermitian Matrix has Real Eigenvalues
Hermitian Matrix has Real Eigenvalues/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf A$ be a Hermitian matrix.
Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its Hermitian conjugate.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$ of $\mathbf A$.
Denote with $\innerprod \cdot \cdot$ the inner product on $\C$. 
We have that $v \ne 0$.
Hence by non-negative definiteness of an inner product:
:$\innerprod v v \ne 0$
and we can divide both sides by $\innerprod v v$.
Thus:
:$\lambda = \overline \lambda$
By Complex Number equals Conjugate iff Wholly Real, $\lambda$ is a real number.
$\lambda$ was arbitrary, so it follows that every eigenvalue is a real number.
Hence the result.
Category:Hermitian Matrix has Real Eigenvalues
Hermitian Operator has Orthogonal Eigenvectors

== Theorem ==
The eigenvectors of a Hermitian operator are orthogonal.
== Proof ==
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the complex numbers $\C$, with a simple spectrum:
:$\hat H \left\vert{x_i}\right\rangle = \lambda_i \left\vert{x_i}\right\rangle$ 
:$\lambda_i \ne \lambda_j$ 
:$\forall i, j \in \N: i \ne j$
Now we compute the following:
and:
From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product:
:$\braket {x_i} {x_j} = \braket {x_j} {x_i}^*$
this becomes:
:$\left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^* = \lambda_j \braket {x_j} {x_i}$
It can be shown that the following relation holds since $\hat H = \hat H^\dagger$:
:$\left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*$
This now gives us the equations:
:$(1): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_i \braket {x_j} {x_i}$
:$(2): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_j \braket {x_j} {x_i}$
Subtracting $(2)$ from $(1)$ gives:
:$\paren {\lambda_i - \lambda_j} \braket {x_j} {x_i} = 0$
Note that $\paren {\lambda_i - \lambda_j} \ne 0$ since we were given $\lambda_i \ne \lambda_j$.
Therefore:
:$\braket {x_j} {x_i} = 0$
Two vectors have inner product $0$ they are orthogonal.
Therefore the eigenvectors of $\hat H$ are orthogonal.
== Source ==
* Hermitian Operator has Real Eigenvalues
Category:Hermitian Operators