nyuuzyou/wikis huggingface dataset:
Linear Transformation is Fredholm Operator iff Pseudoinverse exists

== Theorem ==
Let $U, V$ be vector spaces.
Let $T: U \to V$ be a linear transformation.
Then $T$ is a Fredholm operator $T$ has a pseudoinverse.
== Proof ==
Recall the definitions:
$S$ and $T$ are pseudoinverse to each other :
:$T \circ S - I_U$ is degenerate
and:
:$S \circ T - I_V$ is degenerate.
$T$ is a Fredholm operator :
:$(1): \quad \map \ker T$ is finite-dimensional
:$(2): \quad$ the quotient space $V / \Img T$ is finite-dimensional.
=== Sufficient Condition ===
Let $T$ have a pseudoinverse $S : V \to U$.
That is, both:
:$D_U := S \circ T - I_U$
and:
:$D_V := T \circ S - I_V$
are degenerate.
In particular:
:$\map \ker T \subseteq \map \ker {S \circ T} = \map \ker {D_U + I_U}$
and:
:$\Img {D_V + I_V} = \Img {T \circ S} \subseteq \Img T$
By Degenerate Linear Operator Plus Identity is Fredholm Operator:
:$\map \dim {\map \ker T} \le \map \dim {\map \ker {D_U + I_U} } < +\infty$
and:
:$\map \dim {V / \Img T} \le \map \dim { V / \Img {D_V + I_V} } < +\infty$
=== Necessary Condition ===
Let $T$ be a Fredholm operator.
By Existence of Complementary Subspace, there are subspaces $U' \subseteq U$ and $V' \subseteq V$ such that:
:$U = \map \ker T \oplus U'$
and:
:$V = \Img T \oplus V'$
Consider the projections on direct summands:
:$P : U \to \map \ker T$
and:
:$Q : V \to V'$
Observe:
:$\Img P = \map \ker T$
and:
:$\Img Q = V' \cong V / \Img T$
where the last 
since the isomorphism follow from First Isomorphism Theorem.
Thus , these are finite-dimensional.
In particular, $P$ and $Q$ are degenerate.
By First Isomorphism Theorem:
:$U' \cong U / \map \ker T \cong \Img T$
In particular:
:$T \restriction_{U'} : U' \to \Img T$
is an isomorphism.
Define $S : V \to U$ by:
:$S u = \begin{cases} \paren {T \restriction_{U'} }^{-1} u &: u \in \Img T \\ 0 &: u \in V' \end{cases}$
Then:
:$S \circ T = I_U - P$
:$T \circ S = I_V - Q$
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Fredholm Operators
Category:Linear Algebra
Category:Functional Analysis
Linear Transformation is Injective iff Kernel Contains Only Zero

== Theorem ==
Let $\mathbf V, \mathbf V'$ be vector spaces, with respective zeroes $\mathbf 0, \mathbf 0'$.
Let $T: \mathbf V \to \mathbf V'$ be a linear transformation.
Then:
:$T$ is injective $\map \ker T = \set {\mathbf 0}$
where:
:$\mathbf 0$ is the zero of the domain of $T$
:$\map \ker T$ is the kernel of $T$.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
That $\mathbf 0 \in \map \ker T$ follows from Kernel of Linear Transformation contains Zero Vector.
That $\map \ker T$ is a singleton follows from the definition of injection.
=== Necessary Condition ===
Let $\map \ker T = \set {\mathbf 0}$.
Consider:
:$\map T {\mathbf x} = \mathbf b$
where $\mathbf b$ is in the codomain of $T$.
Let this equation have a solution:
:$\mathbf x = \mathbf x_1 \in \mathbf V$
Suppose $\mathbf x = \mathbf x_2 \in \mathbf V$ is also a solution.
Clearly:
:$\map T {\mathbf x_1} = \map T {\mathbf x_2}$
Observe that:
As $\mathbf x_1, \mathbf x_2$ were arbitrary:
:$\forall \mathbf x_1,\mathbf x_2 \in \mathbf V: \map T {\mathbf x_1} = \map T {\mathbf x_2} \implies \mathbf x_1 = \mathbf x_2$
and the result follows from the definition of injectivity.
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
Category:Linear Transformations
Linear Transformation is Injective iff Kernel Contains Only Zero/Corollary

== Corollary to Linear Transformation is Injective iff Kernel Contains Only Zero ==
Let $\mathbf A$ be in the matrix space $\map {\mathbf M_{m, n} } \R$
Then the mapping: 
:$\R^n \to \R^m: \mathbf x \mapsto \mathbf {A x}$
is injective :
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0}$
where $\map {\mathrm N} {\mathbf A}$ is the null space of $\mathbf A$.
== Proof ==
From Matrix Product as Linear Transformation, $\mathbf x \mapsto \mathbf {A x}$ defines a linear transformation.
The result follows from Linear Transformation is Injective iff Kernel Contains Only Zero and the definition of null space.
== Also see ==
* Null Space Contains Only Zero Vector iff Columns are Independent
== Sources ==
* 
Category:Linear Transformations
Category:Null Spaces
Linear Transformation is Isomorphism iff Inverse Equals Adjoint

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $U : \HH \to \KK$ be a bounded linear transformation.
Then the following are equivalent:
:$(1): \quad U$ is an isomorphism
:$(2): \quad U$ is invertible and $U^{-1} = U^*$, where $U^*$ denotes the adjoint of $U$.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that:
:$U$ is an isomorphism.
That is:
:$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$
for each $g, h \in \HH$.
Let $g, h \in \HH$.
From the definition of the adjoint, we have: 
:$\innerprod g h_\HH = \innerprod g {U^* U h}_\KK$
So, from Inner Product is Sesquilinear, we have: 
:$\innerprod g {h - U^* U h}_\HH = 0$
Setting $g = h - U^* U h$, we have: 
:$\innerprod {h - U^* U h} {h - U^* U h}_\HH = 0$
So from the positiveness of the inner product, we have: 
:$h - U^* U h = 0$
so:
:$h = U^* U h$
for all $h \in \HH$.
So: 
:$U^* U = I_\HH$
where $I_\HH$ is the identity map on $\HH$. 
From Hilbert Space Isomorphism is Bijection, we have that: 
:$U$ is a bijection. 
So: 
:$U$ is invertible.
Let $U^{-1} : \KK \to \HH$ be the inverse of $U$. 
Then, we have: 
So:
:$U$ is invertible with $U^{-1} = U^*$.
=== $(2)$ implies $(1)$ ===
Suppose that: 
:$U$ is invertible with $U^{-1} = U^*$.
By the definition of an isomorphism, we aim to show that: 
:$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$
for each $g, h \in \HH$. 
Let $g, h \in \HH$. 
Then, we have: 
So:
:$U$ is an isomorphism.
== Sources ==
* $\text {II}.2.5$
Category:Linear Transformations on Hilbert Spaces
Category:Adjoints
Linear Transformation of Arithmetic Mean

== Theorem ==
Let $D = \set {x_0, x_1, x_2, \ldots, x_n}$ be a set of real data describing a quantitative variable.
Let $\overline x$ be the arithmetic mean of the data in $D$.
Let $T: \R \to \R$ be a linear transformation such that:
:$\forall i \in \set {0, 1, \ldots, n}: \map T {x_i} = \lambda x_i + \gamma$
Let $T \sqbrk D$ be the image of $D$ under $T$.
Then the arithmetic mean of the data in $T \sqbrk D$ is given by:
:$\map T {\overline x} = \lambda \overline x + \gamma$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5.1$
Category:Arithmetic Mean
Category:Linear Transformations
Category:Linear Transformation of Arithmetic Mean
Linear Transformation of Continuous Random Variable is Continuous Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $a$ be a non-zero real number. 
Let $b$ be a real number. 
Let $X$ be a continuous real variable.
Let $F_X$ be the cumulative distribution function of $X$. 
Then $a X + b$ is a continuous real variable.
Further, if $a > 0$, the cumulative distribution function of $a X + b$, $F_{a X + b}$. is given by: 
:$\ds \map {F_{a X + b} } x = \map {F_X} {\frac {x - b} a}$
for each $x \in \R$.
If $a < 0$, $F_{a X + b}$ is given by: 
:$\ds \map {F_{a X + b} } x = 1 - \map {F_X} {\frac {x - b} a}$
for each $x \in \R$. 
== Proof ==
From Linear Transformation of Real-Valued Random Variable is Real-Valued Random Variable, $a X + b$ is a real-valued random variable.
Since $X$ is a continuous real variable, we have that: 
:$F_X$ is continuous.
We use this fact to show that $F_{a X + b}$ is continuous, showing that $a X + b$ is a continuous real variable.
We split up into cases.
Suppose that $a > 0$. 
Then, for each $x \in \R$, we have: 
From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have: 
:$F_{a X + b}$ is continuous
in the case $a > 0$.
Now suppose that $a < 0$. 
Then, for each $x \in \R$, we have: 
From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have: 
:$F_{a X + b}$ is continuous
in the case $a < 0$.
Category:Continuous Random Variables
Category:Cumulative Distribution Functions
Linear Transformation of Generated Module

== Theorem ==
Let $R$ be a ring.
Let $G$ and $H$ be $R$-modules.
Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$. 
Let $S$ be a generator for $G$.
Suppose that:
:$\forall x \in S: \map \phi x = \map \psi x$
Then $\phi = \psi$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Homomorphism of Generated Group
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.3$
Category:Linear Transformation of Generated Module
Category:Linear Transformations
Linear Transformation of Generated Module/Proof 1

== Theorem ==
== Proof ==
By definitions of generator of module and generated submodule, it follows that:
:$\ds G := \bigcap \set { M' \subseteq G : S \subseteq M', \textrm {$M'$ is a submodule of $G$} }$
Set $S' := \set {x \in G: \map \phi x = \map \psi x}$.
From Elements of Module with Equal Images under Linear Transformations form Submodule, it follows that $S'$ is a submodule of $G$.
By hypothesis, $S \subseteq S'$.
It follows that $S'$ is one of the submodules $M'$.
From Intersection is Subset:Family of Sets, it follows that $G \subseteq S'$.
As $S' \subseteq G$, it follows that $S' = G$.
It follows that for all $x \in G$, we have:
:$\map \phi x = \map \psi x$
which proves that $\phi = \psi$.
Category:Linear Transformation of Generated Module
Linear Transformation of Generated Module/Proof 2

== Theorem ==
== Proof ==
This proof assumes that $R$ is a ring with unity, so $G$ and $H$ become unitary modules.
Let $y \in G$ be arbitrary.
Then by definition of generator, $y$ is the linear combination of elements of $S$:
:$\ds y = \sum_{k \mathop = 1}^n \lambda_k a_k$
for $a_1, a_2, \ldots, a_n \in S, \lambda_1, \lambda_2, \ldots, \lambda n \in R$
Then:
Category:Linear Transformation of Generated Module
Linear Transformation of Normal Random Variable

== Theorem == 
Let $\mu$, $\alpha$ and $\beta$ be real numbers. 
Let $\sigma$ be a positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then: 
:$\alpha X + \beta \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$
 
== Proof ==
Let $Z = \alpha X + \beta$. 
Let $M_Z$ be the moment generating function of $Z$. 
We aim to show that: 
:$Z \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$
By Moment Generating Function of Normal Distribution and Moment Generating Function is Unique, it is sufficient to show that:
:$\map {M_Z} t = \map \exp {\paren {\alpha \mu + \beta} t + \dfrac 1 2 \alpha^2 \sigma^2 t^2}$
We also have, by Moment Generating Function of Normal Distribution, that the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
We have:
Category:Normal Distribution
Linear Transformation of Real-Valued Random Variable is Real-Valued Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Let $a$ and $b$ be real numbers.
Then: 
:$a X + b$ is a real-valued random variable.
== Proof ==
From the definition of a real-valued random variable, we have: 
:$X$ is $\Sigma$-measurable.
We want to verify that $a X + b : \Omega \to \R$ is a $\Sigma$-measurable function.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$a X$ is $\Sigma$-measurable.
From Constant Function is Measurable, we have: 
:$x \mapsto b$ is $\Sigma$-measurable.
From Pointwise Sum of Measurable Functions is Measurable, we have: 
:$a X + b$ is $\Sigma$-measurable.
So:
:$a X + b$ is a real-valued random variable.
Category:Random Variables
Linear Transformation of Submodule

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.
Let $\phi: G \to H$ be a linear transformation.
=== Image of Submodule under Linear Transformation is Submodule ===
=== Preimage of Submodule under Linear Transformation is Submodule ===
=== Image of Linear Transformation is Submodule ===
=== Kernel of Linear Transformation is Submodule ===
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.2$
Category:Linear Transformations
Linear Transformation of Vector Space Equivalent Statements

== Theorem ==
Let $G$ and $H$ be $n$-dimensional vector spaces.
Let $\phi: G \to H$ be a linear transformation.
:$(1): \quad \phi$ is an isomorphism.
:$(2): \quad \phi$ is a monomorphism.
:$(3): \quad \phi$ is an epimorphism.
:$(4): \quad \map \phi B$ is a basis of $H$ for every basis $B$ of $G$.
:$(5): \quad \map \phi B$ is a basis of $H$ for some basis $B$ of $G$.
== Proof ==
$(1)$ implies $(2)$ by definition.
$(2)$ implies $(4)$ by Linear Transformation of Vector Space Monomorphism and Results concerning Generators and Bases of Vector Spaces.
$(4)$ implies $(5)$ by basic logic.
Suppose $\map \phi B$ is a basis of $H$.
Then the image of $\phi$ is a subspace of $H$ generating $H$ and hence is $H$ itself.
Thus $(5)$ implies $(3)$.
Finally, $(3)$ implies that $\phi$ is injective.
If $\phi$ is surjective, the dimension of its kernel is $0$ by Rank Plus Nullity Theorem.
Hence $\phi$ is an isomorphism and therefore $(3)$ implies $(1)$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.7$
Category:Linear Algebra
Linear Transformation of Vector Space Monomorphism

== Theorem ==
Let $G$ and $H$ be a $K$-vector space.
Let $\phi: G \to H$ be a linear transformation.
Then $\phi$ is a monomorphism for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.
== Proof ==
Suppose $\phi$ is a monomorphism.
Let $\sequence {a_n}$ be a linearly independent sequence.
Let:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$
So :
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k = 0$
Hence:
:$\forall k \in \closedint 1 n: \lambda_k = 0$
Suppose that for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.
Let $\map \phi {a_1} = 0$.
Then $a_1 = 0$, otherwise the sequence $\sequence {a_1}$ of one term would be linearly independent but $\sequence {\map \phi {a_1} }$ would not.
Thus $\map \ker \phi = \set 0$ and by the Quotient Theorem for Group Epimorphisms $\phi$ is an isomorphism and therefore a monomorphism.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.6$
Category:Linear Transformations
Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous

== Theorem ==
Linear transformations between finite-dimensional normed vector spaces are continuous. 
== Proof ==
We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.
Choose the Euclidean norm.
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.
Let the matrix $A \in \R^{m \times n}$ be given by:
:$A = \begin {bmatrix}
a_{1 1} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots \\
a_{m 1} & \cdots & a_{m n} \\
\end{bmatrix}$
We have that Set of Linear Transformations is Isomorphic to Matrix Space.
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Then:
Hence:
:$\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$
By Continuity of Linear Transformation between Normed Vector Spaces $T_A$ is continuous.
Hence, all linear transformations between finite-dimensional normed vector spaces are continuous. 
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Linear Transformations
Linearity of Function defined using Function with Translation Property

== Theorem ==
Let $f$ be a real function.
Let $f$ have the translation property.
Let $x$ and $l$ be real numbers.
Define:
:$\map {f_x} l = \map f {x + l} - \map f x$
Then:
:$\forall q \in \Q: \map {f_x} {q l} = q \map {f_x} l$
== Proof ==
=== Lemma ===
 
Let $q$ be a rational number.
Choose integers $n$, $m$ such that:
:$\dfrac n m = q$
We need to prove that:
:$\map {f_x} {q l} = q \map {f_x} l$
We have:
Category:Linear Real Functions
Linearity of Function defined using Function with Translation Property/Lemma

== Lemma ==
Let $f$ be a real function.
Let $f$ have the translation property.
Let $x$ and $l$ be real numbers.
Define:
:$\map {f_x} l = \map f {x + l} - \map f x$
Then:
:$\forall n \in \Z: \map {f_x} {n l} = n \map {f_x} l$
== Proof ==
Let $n$ be an integer.
We have:
Category:Linear Real Functions
Linearity of Function with Translation Property

== Theorem ==
Let $f$ be a real function.
Let $c$ be a real number.
Then:
:$f$ has the translation property
:$\map {f'} c$ exists
:
:$f$ is linear
== Proof ==
=== Sufficient Condition ===
Let:
:$f$ have the translation property
:$\map {f'} c$ exist
We need to show that:
:$f$ is linear
The fact that $f$ has the translation property means:
:$\forall x_1, x_2, t \in \R: \map f {x_1 + t} - \map f {x_2 + t} = \map f {x_1} - \map f {x_2}$
$f$ being linear means:
:$\forall x \in \R: \map f x = ax + b$
where $a$ and $b$ are real numbers.
We have that $f'$ exists by Differentiability of Function with Translation Property.
Also, $f'$ is a constant function by the same theorem.
Accordingly:
:$\map {f'} x = \map {f'} c$ for every real number $x$
Let $x$ be a real number.
We have:
So, $f$ is linear.
=== Necessary Condition ===
Let:
:$f$ be linear
We need to show that:
:$f$ has the translation property
:$\map {f'} c$ exist
The linearity of $f$ means that:
:$\forall x \in \R: \map f x = ax + b$
where $a$ and $b$ are real numbers.
$f$ having the translation property means that:
:$\forall x_1, x_2, t \in \R: \map f {x_1 + t} - \map f {x_2 + t} = \map f {x_1} - \map f {x_2}$
Let $x_1, x_2, t$ be real numbers.
We have:
So, $f$ has the translation property.
It remains to prove that $\map {f'} c$ exists.
We have:
This successful calculation of $\map {f'} c$ demonstrates the existence of $\map {f'} c$.
Category:Linear Real Functions
Linearly Dependent Sequence of Vector Space

== Theorem ==
Let $\struct {G, +}$ be a group whose identity is $\mathbf 0$.
Let $\struct {G, +, \circ}_K$ be a $K$-vector space.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct non-zero vectors of $G$.
Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent :
:$\exists p \in \closedint 2 n: a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$
== Proof ==
=== Necessary Condition ===
Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent.
We have that the set of all integers $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is linearly dependent is not empty.
Let $p$ be its smallest element.
Then from Singleton is Linearly Independent‎, $p \ge 2$, as $a_1 \ne \mathbf 0$ and hence $\set {a_1}$ is linearly independent.
Also, there exist scalars $\lambda_1, \ldots, \lambda_p$, not all of which are zero, such that $\ds \sum_{k \mathop = 1}^p \lambda_k \circ a_k = \mathbf 0$.
 $\lambda_p = 0$.
Then not all of $\lambda_1, \ldots, \lambda_{p-1}$ can be zero.
Then $\sequence {a_k}_{1 \mathop \le k \mathop \le p-1}$ is linearly dependent.
That contradicts the definition of $p$, so $\lambda_p \ne 0$.
So, because:
:$\ds \lambda_p \circ a_p = - \sum_{k \mathop = 1}^{p - 1} \lambda_k \circ a_k$
we must have:
:$\ds a_p = \sum_{k \mathop = 1}^{p - 1} \paren {-{\lambda_p}^{-1} \lambda_k} \circ a_k$
and thus $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.
=== Sufficient Condition ===
Now suppose that $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.
Then:
:$\ds a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k \circ a_k$
So we can assign values to $\lambda_k$ as follows:
:$\forall k \in \closedint 1 n: \lambda_k = \begin{cases}
\mu_k & : k < p \\
-1 & : k = p \\
0 & : k > p \\
\end{cases}$
Then:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = \mathbf 0$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.6$
* : Chapter $7$: Vector Spaces: $\S 33$. Definition of a Basis
* 
Category:Linear Dependence
Category:Vector Spaces
Linearly Independent Set is Basis iff of Same Cardinality as Dimension

== Theorem ==
Let $E$ be a vector space of $n$ dimensions.
Let $H$ be a linearly independent subset of $E$.
$H$ is a basis for $E$ it contains exactly $n$ elements.
== Proof ==
By hypothesis, let $H$ be a linearly independent subset of $E$
=== Necessary Condition ===
Let $H$ be a basis for $E$.
By definition of dimension of vector space, a basis for $E$ contains exactly $n$ elements.
By Bases of Finitely Generated Vector Space have Equal Cardinality, it follows that $H$ also contains exactly $n$ elements.
=== Sufficient Condition ===
Let $H$ contain exactly $n$ elements.
By Sufficient Conditions for Basis of Finite Dimensional Vector Space $H$ is itself a basis for $E$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(v)}$
Category:Bases of Vector Spaces
Category:Dimension of Vector Space
Linearly Independent Set is Contained in some Basis/Finite Dimensional Case

== Theorem ==
Let $E$ be a vector space of $n$ dimensions.
Let $H$ be a linearly independent subset of $E$.
There exists a basis $B$ for $E$ such that $H \subseteq B$.
== Proof 1 ==
== Proof 2 ==
Category:Linearly Independent Set is Contained in some Basis
Linearly Independent Set is Contained in some Basis/Finite Dimensional Case/Proof 1

== Theorem ==
== Proof ==
By hypothesis there is a basis $B$ of $E$ with $n$ elements.
Then $H \cup B$ is a generator for $E$.
So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $E$ such that $H \subseteq C \subseteq H \cup B$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
Category:Linearly Independent Set is Contained in some Basis
Linearly Independent Set is Contained in some Basis/Finite Dimensional Case/Proof 2

== Theorem ==
== Proof ==
Let $H = \set {\xi_1, \xi_2, \ldots, \xi_r}$.
Consider the basis $B = \set {\alpha_1, \alpha_2, \ldots, \alpha_n}$ of $E$.
Consider the set $G = H \cup B = \set {\xi_1, \xi_2, \ldots, \xi_r, \alpha_1, \alpha_2, \ldots, \alpha_n}$.
We have that $G$ is a generator of $E$.
As $B$ is a basis, it follows that each of $H$ is a linear combination of $B$.
Thus $G = H \cup B$ is linearly dependent.
Thus one of the elements $\alpha_i$ of $B$ is a linear combination of the preceding elements of $G$.
Eliminate this one, and do the same thing with with $G \setminus \set {\alpha_i}$. 
Eventually there will exist a set which is a basis of $E$ containing all the elements of $H$.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(i)}$
Category:Linearly Independent Set is Contained in some Basis