nyuuzyou/wikis huggingface dataset:
Zero Matrix is Zero for Matrix Multiplication

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\mathbf A$ be a matrix over $R$ of order $m \times n$
Let $\mathbf 0$ be a zero matrix whose order is such that either:
:$\mathbf {0 A}$ is defined
or:
:$\mathbf {A 0}$ is defined
or both.
Then:
:$\mathbf {0 A} = \mathbf 0$
or:
:$\mathbf {A 0} = \mathbf 0$
whenever they are defined.
The order of $\mathbf 0$ will be according to the orders of the factor matrices.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n}$ be matrices.
Let $\mathbf {0 A}$ be defined.
Then $\mathbf 0$ is of order $r \times m$ for $r \in \Z_{>0}$.
Thus we have:
 
Hence $\mathbf {0 A}$ is the Zero Matrix of order $r \times n$.
Let $\mathbf {A 0}$ be defined.
Then $\mathbf 0$ is of order $n \times s$ for $s \in \Z_{>0}$.
Thus we have:
 
Hence $\mathbf {A 0}$ is the Zero Matrix of order $m \times s$.
If $\mathbf 0$ is of order $n \times m$,then both $\mathbf {A 0}$ and $\mathbf {0 A}$ are defined, and:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $9$
Category:Conventional Matrix Multiplication
Category:Zero Matrix
Zero Measure is Absolutely Continuous with respect to Every Measure

== Definition ==
Let $\struct {M, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {M, \Sigma}$.
Let $\nu$ be the null measure on $\struct {M, \Sigma}$.
Then $\nu$ is absolutely continuous with respect to $\mu$.
== Proof ==
Let $E \in \Sigma$ be such that $\map \mu E = 0$.
By the definition of the null measure, we have:
:$\map \nu E = 0$.
So whenever $E \in \Sigma$ is such that $\map \mu E = 0$, we have $\map \nu E = 0$. 
So $\nu$ is absolutely continuous with respect to $\mu$.
== Sources ==
* 
Category:Absolutely Continuous Measures
Zero Morphism does not Depend on Zero Object

== Theorem ==
Let $\mathbf C$ be a category.
Let $A$ and $B$ be objects of $\mathbf C$.
Let $0_1$ and $0_2$ be zero objects of $\mathbf C$.
Then the morphism defined as the composition 
:$\beta \circ \alpha : A \to 0_1 \to B$ 
of the unique morphism $\alpha : A \to 0_1$ and the unique morphism $\beta : 0_1 \to B$ is equal to the morphism defined as the composition 
:$\delta \circ \gamma : A \to 0_2 \to B$ 
of the unique morphism $\gamma : A \to 0_2$ and the unique morphism $\delta : 0_2 \to B$.
== Proof ==
There are unique morphisms $\epsilon : 0_1 \to 0_2$ and $\zeta : 0_2 \to 0_1$.
Since $0_1$ is terminal, we have
: $\zeta \circ \epsilon = \operatorname{id}_{0_1}$
: $\beta \circ \zeta = \delta$
Since $0_2$ is terminal, we have
: $\epsilon \circ \alpha = \gamma$
Hence
== Notation ==
This justifies the notation $0 := \beta \circ \alpha$, whenever a zero object exists in $\mathbf C$.
Zero Padded Basis Representation

== Theorem ==
Let $b \in \Z: b > 1$.
Let $m \in \Z_{> 0}$.
For every $n \in \Z_{\ge 0}$ such that $n < b^m$, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ such that:
:$(1): \quad \ds n = \sum_{j \mathop = 0}^{m - 1} r_j b^j$
:$(2): \quad \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$
== Informal Proof ==
== Proof ==
Category:Number Bases
Category:Zero Padded Basis Representation
Zero Padded Basis Representation/Proof

== Theorem ==
== Proof ==
=== Case 1 ===
Let $n \in \Z_{> 0} : n < b^m$.
From Basis Representation Theorem there exists a sequence $\sequence {s_j}_{0 \mathop \le j \mathop \le k}$ such that:
:$(a): \quad \ds n = \sum_{j \mathop = 0}^k s_j b^j$
:$(b): \quad \forall j \in \closedint 0 k: s_j \in \N_b$
:$(c): \quad s_k \ne 0$
First it is shown that:
:$k \le m - 1$
:
:$m - 1 < k$
Then:
This contradicts the premise:
:$n < b^m$
It follows that:
:$k \le m - 1$
Let $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ be the sequence defined by:
:$r_j = \begin {cases}
s_j & : \text{if } j \le k\\
0 & : \text{if } k < j \le m - 1
\end {cases}$
Then:
To show that $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is unique, let $\sequence {r'_j}_{0 \mathop \le j \mathop \le m - 1}$ be a sequence such that:
:$(1): \quad \ds n = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$
:$(2): \quad \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$
Since $n > 0$, then: 
:$\exists j \in \closedint 0 {m - 1} : r'_j > 0$ 
Let: 
:$k' = \max \set {j : j \in \closedint 0 {m - 1} \text{ and } r'_j > 0}$
Then:
:$(1): \quad \ds n = \sum_{j \mathop = 0}^{k'} r'_j b^j$
:$(2): \quad \ds \forall j \in \closedint 0 {k'}: r'_j \in \N_b$
:$(3): \quad r_{k'} \ne 0$
and
:$(4): \quad \ds \forall j \in \closedint {k + 1} {m - 1}: r'_j = 0$
By definition the sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le k'}$ is a basis representation for $n$.
From Basis Representation Theorem the basis representation for $n$ is unique.
Then:
:$\sequence {r'_j}_{0 \mathop \le j \mathop \le k'} = \sequence {s_j}_{0 \mathop \le j \mathop \le k} = \sequence {r_j}_{0 \mathop \le j \mathop \le k}$
That is:
:$k' = k$
and
:$\forall j \in \closedint 0 k :r'_j = r_j$ 
It follows that:
:$\forall j \in \closedint {k + 1} {m - 1}: r'_j = 0 = r_j$
So: 
:$\forall j \in \closedint 0 {m - 1}: r'_j = r_j$
This establishes that the sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is unique.
=== Case 2 ===
Let $n = 0$.
Let $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ be the sequence defined by:
:$\forall j \in \closedint 0 {m - 1} : r_j = 0$.
Then:
:$\ds \sum_{j \mathop = 0}^{m - 1} r_j b^j = \sum_{j \mathop = 0}^{m - 1} 0 \cdot b^j = 0$
This shows that $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is a sequence such that:
:$(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$
:$(2): \quad \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$
Let $\sequence {r'_j}_{0 \mathop \le j \mathop \le m - 1}$ be a sequence such that:
:$(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$
:$(2): \quad \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$
Suppose for some $i \in \closedint 0 {m - 1}$, $r'_i > 0$.
Then:
:$\ds \sum_{j \mathop = 0}^{m - 1} r'_j b^j \ge r'_i b^i > 0$.
It follows that:
:$\forall j \in \closedint 0 {m - 1} : r'_j = 0$.
This proves that $0$ has one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ such that:
:$(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$
:$(2): \quad \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$
Category:Zero Padded Basis Representation
Zero Product of Numbers implies Factors are Zero

== Theorem ==
On all the number systems:
:natural numbers $\N$
:integers $\Z$
:rational numbers $\Q$
:real numbers $\R$
:complex numbers $\C$
the following holds.
Let $a \times b = 0$.
Then either $a = 0$ or $b = 0$.
== Proof ==
From Natural Numbers have No Proper Zero Divisors
:$\forall a, b \in \N: a \times b = 0 \implies a = 0 \text { or } b = 0$
We have:
:Integers form Integral Domain
:Rational Numbers form Integral Domain
:Real Numbers form Integral Domain
:Complex Numbers form Integral Domain
Hence by definition of integral domain:
:$a \times b = 0 \implies a = 0 \text { or } b = 0$
where $a, b \in \Z, \Q, \R, \C$.
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
Category:Zero Divisors
Category:Numbers
Zero Product with Proper Zero Divisor is with Zero Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $x \in R$ be a proper zero divisor of $R$.
Then:
:$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$
That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must itself be a zero divisor.
== Proof ==
Follows directly from the definition of proper zero divisor.
If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$.
Category:Zero Divisors
Zero Set of Semi-Inner Product is Vector Subspace

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {V, \innerprod \cdot \cdot}$ be a semi-inner product space over $\GF$.
Let:
:$N = \set {x \in V : \innerprod x x = 0}$
Then $N$ is a vector subspace of $V$.
== Proof ==
Let $x, y \in N$ and 
Then by Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces we have:
:$\cmod {\innerprod x y}^2 \le \innerprod x x \innerprod y y = 0$
and:
:$\cmod {\innerprod y x}^2 \le \innerprod x x \innerprod y y = 0$ 
Hence $\innerprod x y = \innerprod y x = 0$.
Then we have:
Category:Semi-Inner Product Spaces
Zero Simple Staircase Integral Condition for Primitive

== Theorem ==
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $\ds \oint_C \map f z \rd z = 0$ for all simple closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$.
== Proof ==
Let $C$ be a closed staircase contour in $D$, not necessarily simple.
If we show that $\ds \oint_C \map f z \rd z = 0$, then the result follows from Zero Staircase Integral Condition for Primitive.
The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parallel with either the real axis or the imaginary axis.
Denote the parameterization of $C$ as $\gamma: \closedint a b \to \C$, where $\closedint a b$ is a closed real interval.
Denote the parameterization of $C_k$ as $\gamma_k: \closedint {a_k} {b_k} \to \C$.
=== Lemma ===
=== Splitting up the Contour ===
The lemma shows that given a staircase contour $C$, we can assume that for $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$.
This means that in order to intersect itself, $C$ must be a concatenation of at least $4$ directed smooth curves.
Now, we prove the main requirement for Zero Staircase Integral Condition for Primitive, that $\ds \oint_C \map f z \rd z = 0$.
The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of.
=== Basis for the Induction ===
For $n = 1$, $C$ can only be a closed staircase contour if $\gamma$ is constant, so:
For $n = 4$, $C$ can only be a closed staircase contour if $C$ is a simple contour.
Then :
:$\ds \oint_C \map f z \rd z = 0$
This is the basis for the induction.
=== Induction Hypothesis ===
For $N \in \N$, if $C$ is a closed staircase contour that is a concatenation of $n$ directed smooth curves with $n \le N$, then:
:$\ds \oint_C \map f z \rd z = 0$
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
Suppose that $C$ is a closed staircase contour that is a concatenation of $n + 1$ directed smooth curves.
If $C$ is a simple contour, the induction hypothesis is true by the original assumption of this theorem.
Otherwise, define $t_0 = a$, and $t_3 = b$.
Define $t_1 \in \closedint a b$ as the infimum of all $t \in \closedint a b$ for which $\gamma$ intersects itself.
Then define $t_2 \in \hointl {t_1} b$ as the infimum of all $t \in \hointl {t_1} b$ for which $\map \gamma t = \map \gamma {t_1}$.
For $k \in \set {1, \ldots, 3}$, define $\tilde C_k$ as the staircase contour with parameterization $\gamma \restriction {\closedint {t_{k - 1} } {t_k} }$.
Then $\tilde C_2$ is a closed staircase contour that is a concatenation of at least $4$ directed smooth curves.
Then both $\tilde C_1 \cup \tilde C_3$ and $\tilde C_2$ are a concatenation of fewer than $n + 1$ directed smooth curves, so:
Category:Complex Contour Integrals
Category:Zero Simple Staircase Integral Condition for Primitive
Zero Simple Staircase Integral Condition for Primitive/Lemma

== Lemma for Zero Simple Staircase Integral Condition for Primitive ==
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $C$ be a closed staircase contour in $D$.
Then there exists a contour $C'$ such that:
:$\ds \oint_C \map f z \rd z = \oint_{C'} \map f z \rd z$
This contour $C'$ has the property that:
:for all $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$.
== Proof ==
The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of.
=== Basis for the Induction ===
Let $C$ consists of just one directed smooth curve.
Then we write:
:$C' = C$
Thus as $k \in \O$, the proposition is vacuously true.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose that:
:if $C$ is a concatenation of $n$ directed smooth curves, we can find a contour $C'$ with the properties above.
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
Let $C$ be a concatenation of $n + 1$ directed smooth curves.
Suppose that there exists $k \in \set {1, \ldots, n}$ such that the intersection of the images of $C_k$ and $C_{k + 1}$ contain more elements than their common end point $\map {\gamma_k} {b_k}$, otherwise we can put $C' = C$.
Then, one of the two line segments must contain the other line segment.
So:
:$\Img {C_k} \cap \Img {C_{k + 1} } = \gamma_k \sqbrk {\closedint {a'} {b_k} } \cup \gamma_{k + 1} \sqbrk {\closedint {a_{k + 1} } {b'} }$
where either $a' = a_k$, or $b' = b_{k + 1}$.
Define $C_k'$ as the contour that is parameterized as a line segment that goes from $\map {\gamma_k} {a_k}$ to $\map {\gamma_k} {a'}$, and then to $\map {\gamma_{k + 1} } {b_{k + 1} }$.
Then, the image of $C_k'$ is equal to the part of the line segments that does not overlap.
Define:
:$\tilde C_k$ as the contour with the parameterization $\gamma_k \restriction_{\closedint {a'} {b_k} }$
and:
:$\tilde C_{k + 1}$ as the contour with the parameterization $\gamma_{k + 1} \restriction_{\closedint {a_{k + 1} } {b'} }$.
Here, $\gamma_k \restriction_{\closedint {a'} {b_k} }$ denotes the restriction of $\gamma_k$ to $\closedint {a'} {b_k}$.
Now $\tilde C_k$ is the reversed contour of $\tilde C_{k + 1}$.
So:
which shows that the contour integral of $f$ along $C$ is equal to the contour integral of $f$ along $C = C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1}$.
This new contour $C$ is a staircase contour which is a concatenation of $n$ directed smooth curves.
So, by the induction hypothesis, there exists a contour $C'$ such that:
:$\ds \oint_{C'} \map f z \rd z = \oint_{C} \map f z \rd z$
where $C'$ has the desired property.
Category:Zero Simple Staircase Integral Condition for Primitive
Zero Staircase Integral Condition for Primitive

== Theorem ==
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $z_0 \in D$.
Suppose that $\ds \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$ defined by:
:$\ds \map F w = \int_{C_w} \map f z \rd z$
where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.
== Proof ==
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$.
If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \paren {-C_w}$ is a closed staircase contour.
Then the definition of $F$ is independent of the choice of contour, as:
We now show that $F$ is the primitive of $f$.
Let $\epsilon \in \R_{>0}$.
By definition of continuity, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} w \subseteq D$, and for all $z \in \map {B_r} w$:
:$\size {\map f z - \map f w} < \dfrac \epsilon 2$
Let $h = x+iy \in \C \setminus \set 0$ with $x, y \in \R$ such that $\size h < r$.
Let $\LL$ be the staircase contour that goes in a horizontal line from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$.
As $w + x, w + h \in \map {B_r} w$, it follows from Open Ball is Convex Set that $\LL$ is a contour in $\map {B_r} w$.
Then $C_w \cup \LL$ is a staircase contour from $z_0$ to $w + h$, so:
From Derivative of Complex Polynomial, it follows that $\dfrac \d {\d z} \map f w z = \map f w$, so:
We can now show that $\map {F'} w = \map f w$, as:
When $h$ tends to $0$, we have $\map {F'} w = \map f w$ by definition of differentiability.
== Sources ==
* : $\S 2.3$
Category:Complex Analysis
Zero Strictly Precedes One

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $0$ be the zero of $S$.
Let $1$ be the one of $S$.
Then:
:$0 \prec 1$
== Proof ==
This follows directly from the definition of $\prec$.
First note that:
:$\forall n \in S: 0 \preceq n$
from the definition of zero.
Next, from the definition of one:
:$0 \ne 1$
Thus:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Naturally Ordered Semigroup
Zero Subspace is Subspace

== Theorem ==
Let $V$ be a vector space over $K$ with zero vector $\mathbf 0$.
The zero subspace $\set {\mathbf 0}$ is a subspace of $V$.
== Proof ==
We use the Two-Step Vector Subspace Test.
$\set {\mathbf 0}$ is not empty, because it contains $\mathbf 0$.
$\set {\mathbf 0}$ is closed under $+$ because:
:$\forall \mathbf x, \mathbf y \in \set {\mathbf 0}, \mathbf x + \mathbf y = \mathbf 0 + \mathbf 0 = \mathbf 0 \in \set {\mathbf 0}$
$\set {\mathbf 0}$ is closed under multiplication because:
:$\forall \lambda \in K, \mathbf x \in \set {\mathbf 0}: \lambda \mathbf x = \lambda \mathbf 0 = \mathbf 0 \in \set {\mathbf 0}$
Hence the result, from the Two-Step Vector Subspace Test.
== Sources ==
* 
Category:Linear Algebra
Zero Vector Scaled is Zero Vector

== Theorem ==
Let $\struct {\mathbf V, +, \circ}_K$ be a vector space over a division ring $K$, as defined by the vector space axioms.
Then:
:$\forall \lambda \in K: \lambda \circ \bszero = \bszero$
where $\bszero \in \mathbf V$ is the zero vector.
== Proof ==
== Also see ==
* Vector Scaled by Zero is Zero Vector
* Vector Product is Zero only if Factor is Zero
* Zero Vector Space Product iff Factor is Zero
== Sources ==
* : $1$. Preliminary ideas; sets, vector spaces: $1.2$. Vector Spaces
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(iv)}$
Category:Zero Vectors
Zero Vector Space Product iff Factor is Zero

== Theorem ==
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms.
Let $\mathbf v \in \mathbf V, \lambda \in F$.
Then:
:$\lambda \circ \mathbf v = \bszero \iff \paren {\lambda = 0_F \lor x = \bszero}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
Category:Zero Vector Space Product iff Factor is Zero
Category:Zero Vectors
Category:Vector Spaces
Zero Vector Space Product iff Factor is Zero/Proof 1

== Theorem ==
== Proof ==
A vector space is a module, so all results about modules also apply to vector spaces.
So from Scalar Product with Identity it follows directly that:
:$\lambda = 0_F \lor \mathbf v = e \implies \lambda \circ \mathbf v = \bszero$
Next, suppose $\lambda \circ \mathbf v = \bszero$ but $\lambda \ne 0_F$.
Then:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (8)$
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(vi)}$
Category:Zero Vector Space Product iff Factor is Zero
Zero Vector has no Direction

== Theorem ==
A zero vector has no direction.
== Proof ==
Let $\mathbf 0$ denote a zero vector.
 $\mathbf 0$ has a direction.
Then $\mathbf 0$ can be represented as an arrow in a real vector space $\R^n$ with a Cartesian frame.
Let $\mathbf 0$ be so embedded.
Thus it consists of a line segment between two points with an initial point $A$ and a terminal point $B$.
The initial point and a terminal point are distinct from each other.
Let these points be identified as:
Hence we have that the length of $\mathbf 0$ is defined as:
:$\norm {\mathbf 0} = \ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - b_i}^2} > 0$
which means that at least one of $a_i - b_i$ is non-zero.
But this contradicts the definition of $\mathbf 0$ being the zero vector.
It follows by Proof by Contradiction that our assumption that $\mathbf 0$ has a direction must be false.
Hence the result.
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* 
Category:Zero Vectors
Zero Vector in *-Algebra is Hermitian

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra.
Let ${\mathbf 0}_A$ be the zero vector of $A$. 
Then we have:
:${\mathbf 0}_A^\ast = {\mathbf 0}_A$
== Proof ==
By $(\text C^\ast 2)$ in the definition of an involution, we have:
:${\mathbf 0}_A^\ast = \paren { {\mathbf 0}_A + {\mathbf 0}_A}^\ast = {\mathbf 0}_A^\ast + {\mathbf 0}_A^\ast$
Hence:
:${\mathbf 0}_A^\ast = {\mathbf 0}_A^\ast - {\mathbf 0}_A^\ast = {\mathbf 0}_A$
Category:Hermitian Elements of *-Algebras
Zero Vector is Linearly Dependent

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then the singleton set $\set e$ consisting of the zero vector is linearly dependent.
== Proof ==
By Scalar Product with Identity we have:
:$\forall \lambda \in R: \lambda \circ e = e$
Hence the result by definition of linearly dependent.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
Category:Linear Dependence
Zero Vector is Orthogonal to All Vectors

== Theorem ==
Let $\mathbf V$ be a vector space of $n$ dimensions.
Let $\bszero \in \mathbf V$ be the zero vector.
Let $\mathbf a \in \mathbf V$ be an arbitrary vector in $\mathbf V$
Then $\bszero$ is orthogonal to $\mathbf a$.
== Proof ==
By definition, $\mathbf a$ and $\mathbf b$ are orthogonal their dot product is zero:
:$\mathbf a \cdot \mathbf b = 0$
We have:
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
By definition of dot product:
:$\ds \mathbf a \cdot \mathbf b = \sum_{k \mathop = 1}^n a_k b_k$
By definition of zero vector:
Hence:
Hence the result.
== Sources ==
* 
Category:Zero Vectors
Category:Orthogonality (Linear Algebra)
Zero Vector is Unique

== Theorem ==
Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms.
Then the zero vector in $\mathbf V$ is unique:
:$\exists! \mathbf 0 \in \mathbf V: \forall \mathbf x \in \mathbf V: \mathbf x + \mathbf 0 = \mathbf x$
== Proof ==
=== Proof of Existence ===
Follows from the vector space axioms.
=== Proof of Uniqueness ===
Let $\mathbf 0$, $\mathbf 0'$ be zero vectors.
Utilizing the vector space axioms:
== Also see ==
* Additive Inverse in Vector Space is Unique
Category:Zero Vectors
Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE

== Theorem ==
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Let $y_1$ and $y_2$ be linearly independent.
Then their Wronskian is either never zero, or zero everywhere on $\closedint a b$.
== Proof ==
Because $y_1$ and $y_2$ are both particular solutions of $(1)$:
That is:
:$\dfrac {\d P} {\d W} + P W = 0$
This is a linear first order ODE.
From Solution to Linear First Order Ordinary Differential Equation:
:$W = C e^{-\int P \rd x}$
The exponential function is never zero:
Therefore:
:$W = 0 \iff C = 0$
and the result follows.
== Sources ==
* : $\S 3.15$: The General Solution of the Homogeneous Equation: Lemma $1$
Category:Homogeneous LSOODEs
Category:Wronskians