avewright/proofwiki-math huggingface dataset:
Theorem: Star of Ideal of *-Algebra is Ideal

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $I$ be an ideal of $A$.
Define:
:$I^\ast = \set {x^\ast : x \in I}$
Then $I^\ast$ is an ideal of $A$.

Proof:
From Star of Vector Subspace of *-Algebra is Vector Subspace, $I^\ast$ is a vector subspace of $A$.
Let $u \in I^\ast$ and $y \in A$.
Then $u = x^\ast$ for some $x \in I$.
Then we have:
:$u y = x^\ast y = \paren {y^\ast x}^\ast$
from $(\text C^\ast 1)$ and $(\text C^\ast 3)$ in the definition of an involution.
Since $x \in I$, we have $y^\ast x \in I$ since $I$ is an ideal.
Hence $u y \in I$.
Similarly we have:
:$y u = y x^\ast = \paren {x y^\ast}^\ast$
from $(\text C^\ast 1)$ and $(\text C^\ast 3)$ in the definition of an involution.
Hence $y u \in I$.
So $I$ is an ideal of $A$.
{{qed}}
Category:*-Algebras
Category:Ideals of Algebras
j6oqdp7ebt613xafs7xv2kg6ik3oszi
Theorem: Intersection of Ideal of *-Algebra with its Star is Self-Adjoint Ideal

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $I$ be an ideal of $A$.
Define:
:$I^\ast = \set {x^\ast : x \in I}$
Then $I \cap I^\ast$ is a self-adjoint ideal.

Proof:
From Star of Ideal of *-Algebra is Ideal, $I^\ast$ is an ideal.
From Intersection of Algebra Ideals is Ideal, $I \cap I^\ast$ is an ideal.
Further, if $x \in I \cap I^\ast$, then $x \in I$ and there exists $y \in I$ such that $x = y^\ast$.
From $(\text C^\ast 1)$ in the definition of an involution, we have $x^\ast = y^{\ast \ast} = y$. 
So $x^\ast \in I$.
Since $x \in I$ we also clearly have $x^\ast \in I^\ast$.
Hence $x^\ast \in I \cap I^\ast$ for all $x \in I \cap I^\ast$.
Hence $I \cap I^\ast$ is self-adjoint.
{{qed}}
Category:*-Algebras
Category:Ideals of Algebras
49lxtnuwdb2pv9487kiryp8dje0mibj
Theorem: Norm of Element of Approximate Identity of C*-Algebra is Less Than or equal to One

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\family {e_\lambda}_{\lambda \in \Lambda}$ be an approximate identity of $A$.
Then for each $\lambda \in \Lambda$ we have:
:$\norm {e_\lambda} \le 1$

Proof:
From the definition of an approximate identity, for each $\lambda \in \Lambda$ we have:
:$\map {\sigma_A} {e_\lambda} \subseteq \closedint 0 1$
So that, from the definition of spectral radius:
:$\map {r_A} {e_\lambda} \le 1$ 
By definition, each $e_\lambda$ is positive and hence Hermitian.
Hence from Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have $\norm {e_\lambda} = \map {r_A} {e_\lambda} \le 1$ for each $\lambda \in \Lambda$.
{{qed}}
Category:Approximate Identities of C*-Algebras
2w4k7a9613b1sanakhsno13iek3gf7s
Theorem: Closed Ideal of C*-Algebra is Self-Adjoint

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $I$ be a closed ideal of $A$.
Then for each $x \in I$, we have $x^\ast \in I$. 
That is, $I$ is self-adjoint.

Proof:
Let $x \in I$.
Since $I$ is an ideal, we have $x^\ast x \in I$. 
From Generalized Polar Decomposition in C*-Algebra, there exists $u \in A$ such that $x = u \paren {\paren {x^\ast x}^{1/2} }^{1/2}$.
From Power of Power of Positive Element of Unital C*-Algebra, we have $x = u \paren {x^\ast x}^{1/4}$. 
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, $\paren {x^\ast x}^{1/4}$ is positive.
Hence we have by $(\text C^\ast 3)$ in the definition of a $\text C^\ast$-algebra:
:$x^\ast = \paren {x^\ast x}^{1/4} u^\ast$
From Real-Valued Continuous Function Vanishing at Zero applied to Hermitian Element of Closed Ideal of Unital C*-Algebra is contained in Ideal, we have $\paren {x^\ast x}^{1/4} \in I$.
Hence since $I$ is an ideal, we have $x^\ast = \paren {x^\ast x}^{1/4} u^\ast \in I$.
Hence if $x \in I$ then $x^\ast \in I$.
So $I$ is self-adjoint.
{{qed}}
Theorem: Mapping is Bijection iff Direct Image Mapping is Bijection

Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is a bijection
{{iff}}
:$f: S \to T$ is also a bijection.

Proof:
Follows immediately from:
* Mapping is Injection iff Direct Image Mapping is Injection
* Mapping is Surjection iff Direct Image Mapping is Surjection
{{qed}}
Category:Bijections
Category:Direct Image Mappings
3wluh86o3werc9punas37r6q77z8pth
Theorem: Figures in Perspective from Point are in Perspective from Line

Let $A$ and $B$ be plane geometric figures.
Let $A$ and $B$ be in perspective from a point.
Then $A$ and $B$ are also in perspective from a line.

Proof:
{{ProofWanted|A consequence of Desargues' Theorem}}
Theorem: Figures in Perspective from Line are in Perspective from Point

Let $A$ and $B$ be plane geometric figures.
Let $A$ and $B$ be in perspective from a line.
Then $A$ and $B$ are also in perspective from a point.

Proof:
{{ProofWanted|A consequence of Desargues' Theorem}}
Theorem: Separable C*-Algebra has Sequential Approximate Identity

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a separable $\text C^\ast$ algebra.
Then there exists a sequence $\family {e_n}_{n \mathop \in \N}$ forming an approximate identity of $A$.

Proof:
Let $\set {a_n : n \in \N}$ be a countable everywhere dense subset of $A$.
From Existence of Approximate Identity of C*-Algebra arising from Dense Ideal, there exists a directed set $\struct {\Lambda, \preceq}$ and net $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ forming an approximate identity of $A$.
The issue is that, at this point, $\Lambda$ may not be countable.
Hence we desire to extract an $\tuple {\le, \preceq}$-increasing sequence $\sequence {\lambda_n}_{n \mathop \in \N}$ in $\Lambda$ such that $\family {e_{\lambda_n} }_{n \mathop \in \N}$ forms an approximate identity of $A$.
Take $n \in \N$. 
For $1 \le j \le n$, we have that $\family {a_j e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $a_j$.
Hence for each $j$ we can find $\mu_{j, n} \in \Lambda$ such that:
:$\norm {a_j - a_j e_\lambda} < 2^{-n}$ for each $\lambda \succeq \mu_{j, n}$.
Hence $\Lambda$ is directed, there exists $\lambda_n \in \Lambda$ such that:
:$\lambda_n \succeq \mu_{j, n}$ for each $1 \le j \le n$.
Then:
:$\norm {a_j - a_j e_\lambda} < 2^{-n}$ for each $\lambda \succeq \lambda_n$.
By construction, for $i \le j$ we have $\lambda_i \preceq \lambda_j$ and hence $e_{\lambda_i} \le_A e_{\lambda_j}$ where $\le_A$ is the canonical preordering.
Hence from Approximate Identity for Everywhere Dense Subset is Approximate Identity of C*-Algebra, it will suffice to show that:
:$\sequence {a_k e_{\lambda_n} }_{n \mathop \in \N}$ converges to $a_k$ for each $k \in \N$.
Now suppose that $\epsilon > 0$ and $k \in \N$. 
Take $N > k$ such that $2^{-N} < \epsilon$.
Then for $n \ge N$ we have $\lambda_n \succeq \lambda_N$ and hence:
:$\norm {a_k - a_k e_{\lambda_n} } < 2^{-N} < \epsilon$ 
Hence the sequence $\sequence {a_k e_{\lambda_n} }_{n \in \N}$ converges to $a_k$ for each $k \in \N$.
Hence by Approximate Identity for Everywhere Dense Subset is Approximate Identity of C*-Algebra, $\sequence {e_{\lambda_n} }_{n \in \N}$ is an approximate identity of $A$.
{{qed}}
Theorem: Spectrum of Locale as Completely Prime Filters is Sober Space

Let $\struct{L, \preceq}$ be a locale.
Let $\map {\operatorname{Sp}} L$ denote the spectrum as completely prime filters of $L$.
Then:
:$\map {\operatorname{Sp}} L$ is a sober space.

Proof:
By definition of spectrum as completely prime filters:
:$\map {\operatorname{Sp}} L = \struct{\map {\operatorname{pt}} L, \set{\Sigma_a : a \in L}}$
where:
:$\map {\operatorname{pt}} L$ denotes the set of points as completely prime filters of $L$
:$\forall a \in L : \Sigma_a = \set{p \in \map {\operatorname{pt}} L : a \in p}$
Let $\Sigma_L = \set{\Sigma_a : a \in L}$.
Let $\Sigma : L \to \powerset {\map {\operatorname{pt}} L}$ be the mapping defined by:
:$\forall a \in L : \map \Sigma a = \Sigma_a$
where $\powerset {\map {\operatorname{pt}} L}$ denotes  the powerset of $\map {\operatorname{pt}} L$.
From Canonical Mapping of Locale to Powerset of Points is Frame Homomorphism:
:$\Sigma: \struct{L, \preceq} \to \struct{\powerset {\map {\operatorname{pt}} L}, \subseteq}$ is a frame homomorphism.
Theorem: Norm of Non-Negative Increasing Continuous Function applied to Positive Element of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be positive.
Let $\map {\sigma_A} x \subseteq \hointr 0 \infty$ be the spectrum of $x$ in $A$.
Let $f : \map {\sigma_A} x \to \hointr 0 \infty$ be an increasing continuous function.
Let $\map f x$ be obtained from the continuous functional calculus.
Then we have:
:$\norm {\map f x} = \map f {\norm x}$

Proof:
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative:
:$\map f x$ is positive.
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {\map f x}
		| r = \sup \set {\cmod \mu : \mu \in \map {\sigma_A} {\map f x} }
		| c = Spectral Radius of Normal Element of C*-Algebra Equal to Norm
}}
{{eqn	| r = \sup \set {\mu : \mu \in \map {\sigma_A} {\map f x} }
		| c = {{Defof|Positive Element of C*-Algebra}}, $\map {\sigma_A} {\map f x} \subseteq \hointr 0 \infty$
}}
{{eqn	| r = \sup \set {\mu : \mu \in f \sqbrk {\map {\sigma_A} x} }
		| c = Spectral Mapping Theorem
}}
{{eqn	| r = \sup \set {\map f \lambda : \lambda \in \map {\sigma_A} x}
}}
{{eqn	| r = \map f {\sup \set {\lambda : \lambda \in \map {\sigma_A} x} }
		| c = since $f$ is increasing
}}
{{eqn	| r = \map f {\sup \set {\cmod \lambda : \lambda \in \map {\sigma_A} x} }
		| c = {{Defof|Positive Element of C*-Algebra}}, $\map {\sigma_A} x \subseteq \hointr 0 \infty$
}}
{{eqn	| r = \map f {\norm x}
		| c = Spectral Radius of Normal Element of C*-Algebra Equal to Norm
}}
{{end-eqn}}
{{qed}}
Category:Positive Elements of C*-Algebras
pcwo5twqtbbvmwms8ef38possxmjh7r
Theorem: Factorization Theorem for C*-Algebra in terms of Bound on Modulus

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $\le_A$ be the canonical preordering of $A$.
Let $a \in A$ be positive.
Let $x \in A$ be such that:
:$x^\ast x \le_A a$
That is:
:$\cmod x^2 \le_A a$
where $\cmod x$ is the modulus of $x$.
Let $0 < \alpha < 1/2$. 
Then there exists $u \in A$ such that:
:$\norm u \le \norm a^{\frac 1 2 - \alpha}$
and:
:$x = u a^\alpha$
where the meaning of $a^\alpha$ in the case of $A$ not unital is to be clarified, being given directly by the continuous functional calculus otherwise.

Proof:
First take $A$ unital.
Theorem: Product of Powers of Positive Element of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $a \in A$ be positive elements.
Let $\alpha, \beta > 0$ be real numbers.
Let $a^\alpha$, $a^\beta$ and $a^{\alpha + \beta}$ be given by the continuous functional calculus.
Then:
:$a^{\alpha + \beta} = a^\alpha a^\beta$

Proof:
Let $\map {\sigma_A} a \subseteq \hointr 0 \infty$ be the spectrum of $a$.
Define $e_\alpha : \map {\sigma_A} a \to \hointr 0 \infty$, $e_\beta : \map {\sigma_A} a \to \hointr 0 \infty$ and $e_{\alpha + \beta} : \map {\sigma_A} a \to \hointr 0 \infty$ by:
:$\map {e_\alpha} t = t^\alpha$
:$\map {e_\beta} t = t^\beta$
:$\map {e_{\alpha + \beta} } t = t^{\alpha + \beta}$
for each $t \in \map {\sigma_A} a$. 
For $t \in \map {\sigma_A} a$ we have:
:$\map {e_\alpha} t \map {e_\beta} t = t^\alpha t^\beta = t^{\alpha + \beta} = \map {e_{\alpha + \beta} } t$
by Product of Powers.
Hence since the continuous functional calculus is an algebra homomorphism, we obtain:
:$\map {e_\alpha} a \map {e_\beta} a = \map {e_{\alpha + \beta} } a$
That is:
:$a^\alpha a^\beta = a^{\alpha + \beta}$
{{qed}}
Category:Positive Elements of C*-Algebras
g2uxyun135vm6nd2gn8tr4wom5r1a9g
Theorem: Power of Power of Positive Element of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a  $\text C^\ast$-algebra.
Let $a \in A$ be positive elements.
Let $\alpha, \beta > 0$ be real numbers.
Let $a^{\alpha \beta}$ be given by the continuous functional calculus for $a$.
Let $\paren {a^\alpha}^\beta$ be given by the continuous functional calculus for $a^\alpha$.
Then:
:$\paren {a^\alpha}^\beta = a^{\alpha \beta}$

Proof:
First take $A$ unital.
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $\map {\sigma_A} {a^\alpha}$ be the spectrum of $a^\alpha$ in $A$.
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, $a^\alpha$ is positive.
Let $\Theta_{a^\alpha}$ be the continuous functional calculus for $a^\alpha$.
Let $\Theta_a$ be the continuous functional calculus for $a$.
Define $e_{\alpha \beta} : \map {\sigma_A} a \to \hointr 0 \infty$ and $e_\alpha : \map {\sigma_A} \to \hointr 0 \infty$ by:
:$\map {e_{\alpha \beta} } t = t^{\alpha \beta}$
and:
:$\map {e_\alpha} t = t^\alpha$
for each $t \in \map {\sigma_A} a$. 
Define $e_\beta : \map {\sigma_A} {a^\alpha} \to \hointr 0 \infty$ by:
:$\map {e_\beta} t = t^\beta$ 
for each $t \in \map {\sigma_A} {a^\alpha}$.
From the Spectral Mapping Theorem, we have:
:$\map {\sigma_A} {a^\alpha} = \set {t^\alpha : t \in \map {\sigma_A} a}$
Hence we have $e_{\alpha \beta} = e_\beta \circ e_\alpha$ by Power of Power.
So:
{{begin-eqn}}
{{eqn	| l = a^{\alpha \beta}
		| r = \map {\Theta_a} {e_{\alpha \beta} }
		| c = {{Defof|Continuous Functional Calculus}}
}}
{{eqn	| r = \map {\Theta_{\map {e_\alpha} a} } {e_\beta}
		| c = Continuous Functional Calculus obeys Composition
}}
{{eqn	| r = \map {\Theta_{a^\alpha} } {e_\beta} 
}}
{{eqn	| r = \paren {a^\alpha}^\beta 
}}
{{end-eqn}}
Now take $A$ to not be unital.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$.
From the definition of the continuous functional for non-unital $\text C^\ast$-algebras, we have:
{{begin-eqn}}
{{eqn	| l = \tuple {a^{\alpha \beta}, 0}
		| r = \tuple {a, 0}^{\alpha \beta} 
}}
{{eqn	| r = \paren {\tuple {a, 0}^\alpha}^\beta
		| c = from the unital case
}}
{{eqn	| r = \paren {\tuple {a^\alpha, 0} }^\beta 
		| c = {{Defof|Continuous Functional Calculus|subdef = Non-Unital Algebra}} 
}}
{{eqn	| r = \tuple {\paren {a^\alpha}^\beta, 0}
		| c = {{Defof|Continuous Functional Calculus|subdef = Non-Unital Algebra}}
}}
{{end-eqn}}
Hence:
:$a^{\alpha \beta} = \paren {a^\alpha}^\beta$
in this case also.
{{qed}}
Category:Positive Elements of C*-Algebras
1iw95zydwyfug9g05w7divm5e4mviuk
Theorem: Real-Valued Continuous Function Vanishing at Zero applied to Hermitian Element of Closed Ideal of Unital C*-Algebra is contained in Ideal

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $I$ be a closed ideal.
Let $x \in I$ be Hermitian.
Let $\map {\sigma_A} x \subseteq \R$ be the spectrum of $x$ in $A$.
Let $K$ be a compact set such that $0 \in K$ and $\map {\sigma_A} x \subseteq K$.  
Let $f : K \to \R$ be a continuous function with $\map f 0 = 0$.
Let $\map f x$ be obtained from the continuous functional calculus, restricting $f$ to $\map {\sigma_A} x$.
Then $\map f x \in I$.

Proof:
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map \CC {K, \R}$. 
From the Weierstrass Approximation Theorem, there exists a sequence of polynomials $\sequence {p_n}_{n \mathop \in \N}$ such that:
:$\norm {p_n - f}_\infty \to 0$ as $n \to \infty$.
We have:
:$\map {p_n} 0 \to \map f 0 = 0$ as $n \to \infty$.
Take:
:$\map {q_n} x = \map {p_n} x - \map {p_n} 0$
We then have $\map {q_n} 0 = 0$ for each $n \in \N$ and $q_n \to f$ in $\struct {\map \CC {\map {\sigma_A} x, \R}, \norm {\, \cdot \,}_\infty}$.
Since $I$ is an ideal, we have $x^n \in I$ for each $n \ge 1$. 
Since $I$ is a vector subspace, we have $\map {q_n} x \in I$ for each $n \in \N$.
Since the continuous functional calculus is an isometry, we have:
:$\norm {\map {q_n} x - \map f x} = \norm {q_n - f}_\infty \to 0$ as $n \to \infty$ 
So $\map {q_n} x \to \map f x$ as $n \to \infty$. 
Hence since $I$ is closed, we have $\map f x \in I$ as $n \to \infty$. 
{{qed}}
Category:Continuous Functional Calculus
Category:Real-Valued Continuous Function Vanishing at Zero applied to Hermitian Element of Closed Ideal of Unital C*-Algebra is contained in Ideal
hxnnsludg7dhovtg0a6rb2ajp6s8kk4
Theorem: Equation of Plane/General Equation

A plane $P$ is the set of all $\tuple {x, y, z} \in \R^3$, where:
:$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$
where $\alpha_1, \alpha_2, \alpha_3, \gamma \in \R$ are given, and not all of $\alpha_1, \alpha_2, \alpha_3$ are zero.

Proof:
Let $\bsalpha$ be a non-zero vector that is perpendicular to $P$.
Let $A$ be a point in $P$ such that $A = k \bsalpha$ for some $k \in \mathbb R$.
Let $X$ be any other point in $P$.
Let $\mathbf x$ be the vector from the origin to $X$.
Thus, the vector to $X$ from $A$ is $\mathbf x - k \bsalpha$.
By the definition of a vector perpendicular to a plane, $\mathbf x - k \bsalpha$ is always perpendicular to $\bsalpha$.
Thus,
{{begin-eqn}}
{{eqn | l = \bsalpha \cdot \paren {\mathbf x - k \bsalpha}
      | r = 0
      | c = Dot Product of Perpendicular Vectors
}}
{{eqn | l = \bsalpha \cdot \mathbf x - k \bsalpha \cdot \bsalpha
      | r = 0
      | c = Dot Product Operator is Bilinear
}}
{{eqn | ll=\leadsto
      | l = \bsalpha \cdot \mathbf x
      | r = k \bsalpha \cdot \bsalpha
}}
{{end-eqn}}
Let the components of $\bsalpha = \paren {\alpha_1, \alpha_2, \alpha_3}$.
Let the components of $\mathbf x = \paren {x, y, z}$.
Let $\gamma = k \bsalpha \cdot \bsalpha$.
Thus,
{{begin-eqn}}
{{eqn | l = \bsalpha \cdot \mathbf x
      | r = k \bsalpha \cdot \bsalpha
}}
{{eqn | ll=\leadsto
      | l = \alpha_1 x + \alpha_2 y + \alpha_3 z
      | r = \gamma
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Plane/Normal Form

Let $P$ be a plane.
Let $\LL$ be the normal to $P$ through the origin.
Let:
:the length of $\LL$ be $p$
:the direction cosines of $\LL$ be $l$, $m$ and $n$.
Then $P$ can be defined by the equation:
:$l x + m y + n z = p$

Proof:
Let $\mathbf u$ be a unit vector that is perpendicular to $P$ and points towards $P$.
Let $A$ be a point in $P$ such that $A = p \mathbf u$ for some $p \in \mathbb R$.
Therefore, $p$ is the distance from the origin to $P$.
Let $X$ be any other point in $P$.
Let $\mathbf x$ be the vector from the origin to $X$.
Thus, the vector to $X$ from $A$ is $\mathbf x - p \mathbf u$.
By the definition of a vector perpendicular to a plane, $\mathbf x - p \mathbf u$ is always perpendicular to $\mathbf u$.
Thus,
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \paren {\mathbf x - p \mathbf u}
      | r = 0
      | c = Dot Product of Perpendicular Vectors
}}
{{eqn | l = \mathbf u \cdot \mathbf x - p\mathbf u \cdot \mathbf u
      | r = 0
      | c = Dot Product Operator is Bilinear
}}
{{eqn | ll=\leadsto
      | l = \mathbf u \cdot \mathbf x
      | r = p
      | c = $\mathbf u$ is a unit vector and Dot Product of Vector with Itself
}}
{{end-eqn}}
Let the components of $\mathbf u = \paren {l, m, n}$.
Due to the Relation between Direction Cosines, since $l^2 + m^2 + n^2 = 1$, the components $\tuple {l, m, n}$ are direction cosines.
Let the components of $\mathbf x = \paren {x, y, z}$.
Thus,
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf x
      | r = p
}}
{{eqn | ll=\leadsto
      | l = l x + m y + n z
      | r = p
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Plane/Intercept Form

Let $P$ be a plane which intercepts the $x$-axis, $y$-axis and $z$-axis respectively at $\tuple {a, 0, 0}$, $\tuple {0, b, 0}$ and $\tuple {0, 0, c}$, where $a b c \ne 0$.
Then $P$ can be described by the equation:
:$\dfrac x a + \dfrac y b + \dfrac y c = 1$

Proof:
We start with the General Equation of Plane:
:$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$
We then plug in the coordinates for each intercept and solve for the corresponding $\alpha$ coefficient.
{{begin-eqn}}
{{eqn | l = \alpha_1 a
      | r = \gamma
      | c = $\set {a, 0, 0} \in P$
}}
{{eqn | ll=\leadsto
      | l = \alpha_1
      | r = \frac \gamma a
}}
{{eqn | l = \alpha_2 b 
      | r = \gamma
      | c = $\set {0, b, 0} \in P$
}}
{{eqn | ll=\leadsto
      | l = \alpha_2 
      | r = \frac \gamma b
}}
{{eqn | l = \alpha_3 c
      | r = \gamma
      | c = $\set {0, 0, c} \in P$
}}
{{eqn | ll= \leadsto
      | l = \alpha_3
      | r = \frac \gamma c
}}
{{end-eqn}}
Substituting the $\alpha$ values into the General Equation of Plane results in
{{begin-eqn}}
{{eqn | l = \frac \gamma a x + \frac \gamma b y + \frac \gamma c z
      | r = \gamma
}}
{{eqn | ll=\leadsto
      | l = \frac x a + \frac y b + \frac z c
      | r = 1
      | c = Divide by $\gamma$
}}
{{end-eqn}}
{{qed}}
Theorem: Three-Point Form of Equation of Plane/Determinant Form

Let $P_1 := \tuple {x_1, y_1, z_1}$, $P_2 := \tuple {x_2, y_2, z_2}$ and $P_3 := \tuple {x_3, y_3, z_3}$ be non-collinear points in a cartesian $3$-space.
Let $P$ be the plane passing through $P_1$, $P_2$ and $P_3$.
$P$ can be expressed in the form:
:$\begin {vmatrix} x & y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \end {vmatrix} = 0$

Proof:
=== Lemma: $\set {P_1, P_2, P_3} \in P$ ===
Substituting any of $\set {\tuple {x_1, y_1, z_1}, \tuple {x_2, y_2, z_2}, \tuple {x_3, y_3, z_3} }$ for $\tuple {x, y, z}$ satisfies the equation due to Square Matrix with Duplicate Rows has Zero Determinant.
Therefore, the set of points defined by the equation contains $P_1$, $P_2$ and $P_3$.
{{qed|lemma}}
Theorem: Three-Point Form of Equation of Plane/Vector Form

Let $P_1$, $P_2$ and $P_3$ be non-collinear points in a cartesian $3$-space whose position vectors are $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Let $P$ be the plane passing through $P_1$, $P_2$ and $P_3$.
$P$ can be expressed in the form:
:$\mathbf r = \mathbf a + \lambda \paren {\mathbf b - \mathbf a} + \mu \paren {\mathbf c - \mathbf a}$
where:
:$\mathbf r$ is the position vector of an arbitrary point on $P$
:$\lambda$ and $\mu$ are real parameters.

Proof:
Let $\bsalpha$ be the non-zero cross product:
:$\bsalpha = \paren {\mathbf b - \mathbf a} \times \paren {\mathbf c - \mathbf a}$
If $P_1$, $P_2$, and $P_3$ were collinear, then $\mathbf b - \mathbf a$ and $\mathbf c - \mathbf a$ would be parallel or antiparallel.
Since both the cross product of parallel vectors and the cross product of antiparallel vectors are the zero vector, $P_1$, $P_2$, and $P_3$ cannot be collinear.
Apply the dot product of $\bsalpha$ to both sides of the equation:
{{begin-eqn}}
{{eqn | l = \mathbf r \cdot \bsalpha
      | r = \paren {\mathbf a  + \lambda \paren {\mathbf b - \mathbf a} + \mu \paren {\mathbf c - \mathbf a} } \cdot \bsalpha
}}
{{eqn | r = \mathbf a \cdot \bsalpha + \lambda \paren {\mathbf b - \mathbf a} \cdot \bsalpha + \mu \paren {\mathbf c - \mathbf a} \cdot \bsalpha
      | c = Dot Product Operator is Bilinear
}}
{{eqn | r = \mathbf a \cdot \bsalpha
      | c = Vector Cross Product is Orthogonal to Factors and Dot Product of Perpendicular Vectors
}}
{{end-eqn}}
Let the components of $r$ be $\tuple {x, y, z}$.
Let the components of $\bsalpha$ be $\tuple {\alpha_1, \alpha_2, \alpha_3}$.
Let $\gamma = \mathbf a \cdot \bsalpha$.
Then,
{{begin-eqn}}
{{eqn | l = \mathbf r \cdot \bsalpha
      | r = \mathbf a \cdot \bsalpha
}}
{{eqn | ll= \leadsto
      | l = \alpha_1 x + \alpha_2 y + \alpha_3 z
      | r = \gamma
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
which is the equation of a plane.
Notice that, if $\bsalpha = \mathbf 0$, the coefficients $\alpha_1$, $\alpha_2$, and $\alpha_3$ would violate the requirement that not all of them are zero.
{{qed}}