nyuuzyou/wikis huggingface dataset:
Linear Transformation Compact iff Adjoint Compact

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $T: H \to K$ be a linear transformation.
Then $T$ is compact its adjoint $T^*$ is.
== Proof ==
First suppose that $T$ is compact.
From the construction of the adjoint, $T^\ast$ is bounded.
From Right Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, $T T^\ast$ is compact.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $K$.
Then there exists a convergent subsequence of $\sequence {T T^\ast x_n}_{n \mathop \in \N}$, say $\sequence {T T^\ast x_{n_j} }_{j \mathop \in \N}$. 
We show that $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ converges.
Pick $M > 0$ such that: 
:$\norm {x_n}_K \le M$
for each $n \in \N$. 
We have: 
Since $\sequence {\map {T T^\ast} {x_{n_j} } }_{j \mathop \in \N}$ is convergent, it is Cauchy from Convergent Sequence is Cauchy Sequence. 
Let $\epsilon > 0$. 
Pick $N \in \N$ so that: 
:$\ds \norm {\map {T T^\ast} {x_{n_j} - x_{n_k} } }_K < \frac {\epsilon^2} {2 M}$
for $n, m \ge N$. 
Then for $n, m \ge N$ we have: 
:$\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K^2 < \epsilon^2$
so:
:$\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K < \epsilon$
So $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is Cauchy.
Since $H$ is Hilbert, $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is convergent.
So for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $K$, $\sequence {T^\ast x_n}_{n \mathop \in \N}$ has a convergent subsequence. 
So $T^\ast$ is compact.
So for a bounded linear transformation $T : H \to K$, if $T$ is compact then $T^\ast$ is compact.
So for a bounded linear transformation $T : H \to K$, if $T^\ast$ is compact then $T^{\ast \ast}$ is compact.
From Adjoint is Involutive, we therefore have that that if $T^\ast$ is compact then $T$ is compact.
So $T$ is compact its adjoint $T^*$ is.
== Sources ==
* $II.4.4$
* $15.3$: Compact Linear Operators
Category:Adjoints
Category:Compact Linear Transformations
Linear Transformation Maps Zero Vector to Zero Vector

== Theorem ==
Let $\mathbf V$ be a vector space, with zero $\mathbf 0$.
Likewise let $\mathbf V\,'$ be another vector space, with zero $\mathbf 0'$.
Let $T: \mathbf V \to \mathbf V\,'$ be a linear transformation.
Then:
:$T: \mathbf 0 \mapsto \mathbf 0'$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Linear Transformations
Category:Linear Transformation Maps Zero Vector to Zero Vector
Linear Transformation as Matrix Product

== Theorem ==
Let $T: \R^n \to \R^m, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.
Then:
:$\map T {\mathbf x} = \mathbf A_T \mathbf x$
where $\mathbf A_T$ is the $m \times n$ matrix defined as:
:$\mathbf A_T = \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatrix}$
where $\tuple {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n}$ is the standard ordered basis of $\R^n$.
== Proof ==
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$.
Let $\mathbf I_n$ be the unit matrix of order $n$.
Then:
That $\mathbf A_T$ is $m \times n$ follows from each $\map T {\mathbf e_i}$ being an element of $\R^m$ and thus having $m$ rows.
== Also see ==
* Matrix Product as Linear Transformation
== Sources ==
* 
* 
* 
Category:Linear Transformations
Linear Transformation between Normed Vector Spaces is Bounded iff Bounded as Linear Transformation between Topological Vector Spaces

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be linear transformations. 
Then $T$ is bounded as a linear transformation between normed vector spaces it is bounded as a linear transformation between topological vector spaces. 
== Proof ==
=== Sufficient Condition ===
Suppose that $T$ is bounded as a linear transformation between topological vector spaces
Then:
:for each von Neumann-bounded subset $E$ of $X$, $T \sqbrk E$ is von Neumann-bounded.
From Characterization of von Neumann-Boundedness in Normed Vector Space, this is equivalent to: 
:if $E \subseteq X$ is such that $\norm x_X < M$ for each $x \in E$ and some $M > 0$, then there exists $M' > 0$ such that $\norm {T x}_Y < M'$ for each $x \in E$.
In particular:
:there exists $M' > 0$ such that $\norm {T x}_Y < M'$ for each $x \in X$ with $\norm x_X < 1$. 
Then for $x \ne \mathbf 0_X$, we have: 
:$\ds \norm {\map T {\frac x {2 \norm x_X} } }_Y < M'$
so that: 
:$\ds \norm {T x}_Y < 2 M' \norm x_X$
for each $x \ne \mathbf 0_X$.
Hence $\norm {T x}_Y \le 2 M' \norm x_X$ for all $x \in X$. 
So $T$ is bounded as a linear transformation between normed vector spaces.
=== Necessary Condition ===
Suppose that bounded as a linear transformation between normed vector spaces.
Again from Characterization of von Neumann-Boundedness in Normed Vector Space, we need to show that: 
:if $E \subseteq X$ is such that $\norm x_X < M$ for each $x \in E$ and some $M > 0$, then there exists $M' > 0$ such that $\norm {T x}_Y < M'$ for each $x \in E$.
Let $E \subseteq X$ be such that $\norm x_X < M$ for each $x \in E$.
From the definition of a bounded linear transformation between normed vector spaces, there exists $K > 0$ such that: 
:$\norm {T x}_Y \le K \norm x_X$
for each $x \in X$. 
Then for $x \in E$ we have $\norm {T x}_Y < M' K$.
So $T$ is bounded as a linear transformation between topological vector spaces. 
Category:Bounded Linear Transformations (Topological Vector Spaces)
Category:Bounded Linear Transformations
Linear Transformation between Normed Vector Spaces is Open iff Image of Open Unit Ball is Open

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$. 
Let $T : X \to Y$ be a linear transformation.
Let $B_X^O$ be the open unit ball of $X$.
Then $T$ is open $T \sqbrk {B_X^O}$ is open.
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is open.
From Open Ball is Open Set, $B_X^O$ is open in $\struct {X, \norm {\, \cdot \,}_X}$.
So $T \sqbrk {B_X^O}$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$.
=== Sufficient Condition ===
Suppose that $T \sqbrk {B_X^O}$ is open.
Let $U$ be an open set in $X$.
From the definition of an open set:
:for each $x \in U$ there exists $\epsilon_x > 0$ such that:
::$x + \epsilon_x B_X^O \subseteq U$
Then we have: 
:$\ds U = \bigcup_{x \in U} \paren {x + \epsilon_x B_X^O}$
Hence:
From Dilation of Open Set in Normed Vector Space is Open:
:$\epsilon_x T \sqbrk {B_X^O}$ is open for each $x \in U$.
From Translation of Open Set in Normed Vector Space is Open
:$T x + \epsilon_x T \sqbrk {B_X^O}$ is open for each $x \in U$.
Hence from the definition of a topology:
:$\ds T \sqbrk U = \bigcup_{x \in U} \paren {T x + \epsilon_x T \sqbrk {B_X^O} }$ is open.
So $T \sqbrk U$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$ whenever $U$ is open in $\struct {X, \norm {\, \cdot \,}_X}$.
So $T$ is an open mapping. 
Category:Linear Transformations
Category:Open Mappings
Category:Normed Vector Spaces
Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin

== Theorem ==
Let $K$ be a topological field.
Let $X$ and $Y$ be topological vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Then $T$ is everywhere continuous it is continuous at ${\mathbf 0}_X$.
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is everywhere continuous.
Then $T$ is continuous at every point.
In particular, $T$ is continuous at ${\mathbf 0}_X$.
=== Sufficient Condition ===
Suppose that $T$ is continuous at ${\mathbf 0}_X$.
We aim to show that $T$ is continuous at every point.
Let $x \in X$.
Let $V$ be an open neighborhood of $T x \in Y$.
We aim to show that there exists an open neighborhood $U$ of $x$ such that: 
:$T \sqbrk U \subseteq V$
From Translation of Open Set in Topological Vector Space is Open, we have that: 
:$V - T x$ is an open neighborhood of ${\mathbf 0}_Y$.
By Linear Transformation Maps Zero Vector to Zero Vector, we have:
:$\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$
Since $T$ is continuous at ${\mathbf 0}_X$, there exists an open neighborhood $U'$ of ${\mathbf 0}_X$ such that: 
:$T \sqbrk {U'} \subseteq V - T x$
Then we have, by translation by $T x$: 
:$T \sqbrk {U'} + T x \subseteq V$
From Image of Translation of Set under Linear Transformation is Translation of Image, we then have: 
:$T \sqbrk {U' + x} \subseteq V$
Since $U'$ is an open neighborhood of ${\mathbf 0}_X$, $U = U' + x$ is an open neighborhood of $x$ with: 
:$T \sqbrk U \subseteq V$ 
Hence the result.
== Sources ==
* : $1.17$: Theorem
Category:Topological Vector Spaces
Category:Linear Transformations
Linear Transformation between Topological Vector Spaces is Open iff Interior of Image of Open Neighborhood contains Zero Vector

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Then $T$ is open :
:for each open neighborhood $U$ of ${\mathbf 0}_X$, we have ${\mathbf 0}_Y \in T \sqbrk U^\circ$.
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is open.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$ in $\struct {X, \tau_X}$.
Since $T$ is open, $T \sqbrk U$ is an open neighborhood of $\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$ in $\struct {Y, \tau_Y}$.
In particular, from the definition of interior, we have ${\mathbf 0}_Y \in T \sqbrk U^\circ$.
=== Sufficient Condition ===
Suppose that:
:for each open neighborhood $U$ of ${\mathbf 0}_X$, we have ${\mathbf 0}_Y \in T \sqbrk U^\circ$.
That is:
:for each open neighborhood $U$ of ${\mathbf 0}_X$, there exists an open neighborhood $O$ of ${\mathbf 0}_Y$ such that $O \subseteq T \sqbrk U$.
Let $V \subseteq X$ be open in $\struct {X, \tau_X}$.
We want to show that $T \sqbrk V \subseteq Y$ is open in $\struct {Y, \tau_Y}$.
Let $x \in V$ so that $T x \in T \sqbrk V$. 
From Image of Translation of Set under Linear Transformation is Translation of Image, $T \sqbrk {V - x}$ is an open neighborhood of ${\mathbf 0}_Y$ in $\struct {Y, \tau_Y}$.
By hypothesis, there exists an open neighborhood $O_x$ of ${\mathbf 0}_X$ such that:
:$O_x \subseteq T \sqbrk {V - x}$
so that:
:$O_x + T x \subseteq T \sqbrk V$ for all $x \in V$.
That is: 
:$\ds \bigcup_{x \mathop \in V} \paren {O_x + T x} \subseteq T \sqbrk V$
Conversely, we have:
:$\ds T \sqbrk V \subseteq \bigcup_{x \mathop \in V} \paren {O_x + T x}$
So, we obtain:
:$\ds T \sqbrk V = \bigcup_{x \mathop \in V} \paren {O_x + T x}$
From Image of Translation of Set under Linear Transformation is Translation of Image, $O_x + T x$ is an open neighborhood of $T x$ in $\struct {Y, \tau_Y}$.
So $T \sqbrk V$ is the union of open sets in $\struct {Y, \tau_Y}$, it is open in $\struct {Y, \tau_Y}$.
Since $V \subseteq X$ was an arbitrary open set in $\struct {X, \tau_X}$, we have that $T$ is an open mapping.
Category:Open Mappings
Category:Topological Vector Spaces
Linear Transformation defined from Graph

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $X \times Y$ be the direct product of $X$ and $Y$.
Let $U$ be a vector subspace of $X \times Y$ such that:
:whenever $\tuple { {\mathbf 0}_X, y} \in U$, we have $y = {\mathbf 0}_Y$
Define:
:$\map D T = \set {x \in X : \tuple {x, y} \in U \text { for some } y \in Y}$ 
Then there exists a linear transformation $T : \map D T \to Y$ with graph $\map \GG T = U$. 
== Proof ==
We first show that $\map D T$ is a vector subspace of $X$.
We have $\tuple { {\mathbf 0}_X, {\mathbf 0}_Y} \in U$, so ${\mathbf 0}_X \in \map D T$. 
Hence $\map D T \ne \O$. 
Let $x, y \in \map D T$ and $\alpha \in \C$.
Then there exists $u, v \in Y$ such that $\tuple {x, u} \in U$ and $\tuple {y, v} \in U$. 
Since $U$ is a vector subspace we have:
:$\tuple {x, u} + \alpha \tuple {y, v} = \tuple {x + \alpha y, u + \alpha v} \in U$
So $x + \alpha y \in \map D T$.
Hence from the One-Step Vector Subspace Test, we have that $\map D T$ is a vector subspace of $X$.
Suppose that $\tuple {x, y_1}, \tuple {x, y_2} \in U$.
Then $\tuple { {\mathbf 0}_X, y_1 - y_2} = \tuple {x, y_1} - \tuple {x, y_2} \in U$ since $U$ is a vector subspace of $X \times Y$. 
So $y_1 - y_2 = {\mathbf 0}_Y$ and hence $y_1 = y_2$. 
Hence we can define $T x$ as the unique $y$ such that $\tuple {x, y} \in U$ for $x \in \map D T$.
Clearly we will have $\map G T = U$. 
We just need to show that $T$ is linear.
Let $x, y \in \map D T$ and $\alpha \in \C$.
Then we have $\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, T x + \alpha T y} \in U$.
Since $\map T {x + \alpha y}$ is unique with $\tuple {x + \alpha y, \map T {x + \alpha y} } \in U$, we have:
:$\map T {x + \alpha y} = T x + \alpha T y$
Hence $T$ is linear.
Category:Linear Transformations
Linear Transformation from Banach *-Algebra is Bounded if Bounded on Hermitian Elements

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ be a Banach $\ast$-algebra.
Let $\struct {B, \norm {\, \cdot \,}_B}$ be a normed vector space over $\C$.
Let $T : A \to B$ be a linear transformation such that there exists $M > 0$ with:
:$\norm {T x}_B \le M \norm x_A$ for each $x \in A$ Hermitian.
Then $T$ is bounded.
== Proof ==
Let $x \in A$.
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, we have:
:$x = \map \Re x + i \map \Im x$
where $\map \Re x$ and $\map \Im x$ are Hermitian.
Then we have:
Hence $T$ is bounded.
Category:Banach *-Algebras
Linear Transformation from C*-Algebra is Bounded if Bounded on Positive Elements

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ be a $\text C^\ast$-algebra.
Let $\struct {B, \norm {\, \cdot \,}_B}$ be a normed vector space over $\C$.
Let $T : A \to B$ be a linear transformation such that there exists $M > 0$ with:
:$\norm {T x}_B \le M \norm x_A$ for each $x \in A$ positive.
Then $T$ is bounded.
== Proof ==
From Linear Transformation from Banach *-Algebra is Bounded if Bounded on Hermitian Elements, it is enough to show that there exists $M' > 0$ such that:
:$\norm {T x}_B \le M' \norm x_A$ for each $x \in A$ Hermitian.
Let $x \in A$ be Hermitian. 
From Hermitian Element of C*-Algebra Decomposes into Positive Elements, there exists positive $x^+, x^-$ such that:
:$x = x^+ - x^-$ 
with:
:$\norm {x^+}_A \le \norm x_A$ and $\norm {x^-}_A \le \norm x_A$
Then we have:
for all $x \in A$. 
Hence $M' = 2 M$ satisfies:
:$\norm {T x}_B \le M' \norm x_A$ for each $x \in A$ Hermitian.
So by Linear Transformation from Banach *-Algebra is Bounded if Bounded on Hermitian Elements, $T$ is bounded.
Category:C*-Algebras
Linear Transformation from Cartesian Space on Hausdorff Topological Field to Topological Vector Space is Continuous

== Theorem ==
Let $K$ be a topological field.
Let $K^n$ be the $n$-Cartesian space as a topological vector space with its product topology.
Let $X$ be a topological vector space over $K$.
Let $f : K^n \to X$ be a linear transformation.
Then $f$ is continuous.
== Proof ==
Let $\set {e_1, \ldots, e_n}$ be the standard basis for $K^n$.
For each $\alpha_1, \alpha_2, \ldots, \alpha_n$ we have: 
:$\ds \map f {\sum_{i \mathop = 1}^n \alpha_i e_i} = \sum_{i \mathop = 1}^n \alpha_i \map f {e_i}$
Let $\pr_i : K^n \to K$ be the projection of $K^n$ onto its $i$th factor, so that: 
:$\ds \map {\pr_i} {\sum_{i \mathop = 1}^n \alpha_i e_i} = \alpha_i$
Then we have: 
:$\ds \map f x = \sum_{i \mathop = 1}^n \map {\pr_i} x \map f {e_i}$ 
for each $x \in K^n$.
From Projection from Product Topology is Continuous:
:$\pr_i$ is continuous for each $1 \le i \le n$. 
From Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous:
:$f$ is continuous.
== Sources ==
* : $1.20$: Lemma
Category:Topological Fields
Category:Topological Vector Spaces
Linear Transformation from Center of Scalar Ring

== Theorem ==
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\phi: G \to H$ be a linear transformation.
Let $\map Z R$ be the center of the scalar ring $R$.
Let $\lambda \in \map Z R$.
Then $\lambda \circ \phi$ is a linear transformation.
== Proof ==
By definition of linear transformation, we need to show that:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
:$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\paren {\lambda \circ \phi} } x$
Let $\lambda \in \map Z R$.
Then:
Because $\lambda \in \map Z R$, $\lambda$ commutes with all elements of $R$.
So:
:$\forall \mu \in R: \lambda \circ \mu = \mu \circ \lambda$.
Thus:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective

== Theorem ==
Let $K$ be a field. 
Let $V$ be a finite dimensional vector space over $K$.
Let $f: V \to V$ be a linear transformation on $V$.
Then $f$ is an injection $f$ is a surjection.
== Proof ==
Let $n = \dim V$. 
From Vector Space has Basis, there exists a basis: 
:$\BB = \set {e_1, \ldots, e_n}$ 
for $V$. 
Then from Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image, $f \sqbrk \BB$ is a generating set for $f \sqbrk V$. 
=== Sufficient Condition ===
Suppose that $f$ is a surjection. 
Then $f \sqbrk V = V$. 
So $f \sqbrk \BB$ is a generating set for $f \sqbrk V$ of size $n$. 
From Sufficient Conditions for Basis of Finite Dimensional Vector Space, $f \sqbrk \BB$ is therefore a basis for $V$. 
In particular, $\map f {e_1}, \map f {e_2}, \ldots, \map f {e_n}$ are linearly independent.
Now take $x \in V$ such that $\map f x = \mathbf 0_V$ and write: 
:$\ds x = \sum_{i \mathop = 1}^n \alpha_i e_i$ 
for coefficients $\alpha_1, \ldots, \alpha_n \in K$. 
Since $f$ is linear, we have: 
:$\ds \mathbf 0_V = \map f x = \sum_{i \mathop = 1}^n \alpha_i \map f {e_i}$ 
Since $\map f {e_1}, \map f {e_2}, \ldots, \map f {e_n}$ are linearly independent, we have $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$. 
Hence $x = 0$. 
So we have $\ker f = \set {\mathbf 0_V}$. 
From Linear Transformation is Injective iff Kernel Contains Only Zero, $f$ is injective.
=== Necessary Condition ===
Suppose that $f$ is injective.
From Linear Transformation is Injective iff Kernel Contains Only Zero, $\ker f = \set {\mathbf 0_V}$. 
So if $\alpha_1, \alpha_2, \ldots, \alpha_n \in K$ are such that: 
Then we have:
:$\ds \sum_{i \mathop = 1}^n \alpha_i e_i = \mathbf 0_V$
and hence $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$ since $e_1, e_2, \ldots, e_n$ is linearly independent. 
So $f \sqbrk \BB$ is a linearly independent set of size $n$. 
So $\map \span {f \sqbrk \BB} = f \sqbrk V$ is a subspace of $V$ of dimension $n$. 
From Dimension of Proper Subspace is Less Than its Superspace, we have $f \sqbrk V = V$. 
So $f$ is surjective.
== Sources ==
* : $5$: Compact spaces: $5.2$: Definition of compactness
Category:Linear Transformations
Linear Transformation from Ordered Basis less Kernel

== Theorem ==
Let $G$ and $H$ be unitary $R$-modules.
Let $\phi: G \to H$ be a non-zero linear transformation.
Let $G$ be $n$-dimensional.
Let $\sequence {a_n}$ be any ordered basis of $G$ such that $\set {a_k: r + 1 \le k \le n}$ is the basis of the kernel of $\phi$.
Then $\sequence {\map \phi {a_r} }$ is an ordered basis of the image of $\phi$.
== Proof ==
Suppose:
:$\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^r \lambda_k a_k} = 0$
So $\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k}$ belongs to the kernel of $\phi$ and hence is also a linear combination of $\set {a_k: r + 1 \le k \le n}$.
Thus $\forall k \in \closedint 1 r: \lambda_k = 0$ since $\sequence {a_n}$ is linearly independent.
Thus the sequence $\sequence {\map \phi {a_r} }$ is linearly independent.
We have $\forall k \in \closedint {r + 1} n: \map \phi {a_k} = 0$.
So let $x \in G$.
Let $\ds x = \sum_{k \mathop = 1}^n \mu_k a_k$.
Then:
:$\ds \map \phi x = \sum_{k \mathop = 1}^n \mu_k \map \phi {a_k} = \sum_{k \mathop = 1}^r \mu_k \map \phi {a_k}$
Therefore $\sequence {\map \phi {a_r} }$ is an ordered basis of the image of $\phi \sqbrk G$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.5$
Category:Linear Transformations
Linear Transformation has Finite Rank iff Domain Quotiented by Kernel is Finite Dimensional

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $T : X \to Y$ be a linear transformation. 
Let $\ker T$ be the kernel of $T$.
Let $X/\ker T$ be the quotient vector space of $X$ modulo $\ker T$.
Then $T$ has finite rank $X/\ker T$ is finite dimensional.
== Proof ==
Let $q : X \to X/\ker T$ be the quotient mapping. 
=== Sufficient Condition ===
Suppose that $X/\ker T$ is finite dimensional.
Let $\set {\map q {x_1}, \map q {x_2}, \ldots, \map q {x_n} }$ be a basis for $X/\ker T$.
Let $x \in X$. 
Then there exists $\alpha_1, \alpha_2, \ldots, \alpha_n \in \GF$ such that: 
:$\ds \map q x = \sum_{i \mathop = 1}^n \alpha_i \map q {x_i}$
From Quotient Mapping is Linear Transformation, we have: 
:$\ds \map q {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$
So there exists $f \in \ker T$ such that: 
:$\ds x = f + \sum_{i \mathop = 1}^n \alpha_i x_i$
Then: 
:$\ds T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$
since $T$ is linear.
So the range of $T$ is generated by $\set {T x_1, T x_2, \ldots, T x_n}$.
=== Necessary Condition ===
Suppose that the range of $T$ is finite dimensional. 
Let: 
:$\set {T x_1, \ldots, T x_n}$ 
be a basis for the range of $T$.
Let $x \in X$. 
Then there exists $\alpha_1, \ldots, \alpha_n \in \GF$ such that: 
:$\ds T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$
Then: 
:$\ds \map T {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$
That is: 
:$\ds x - \sum_{i \mathop = 1}^n \alpha_i x_i \in \ker T$ 
so that: 
:$\ds \map q {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$ 
Then: 
:$\ds \map q x = \sum_{i \mathop = 1}^n \alpha_i \map q {x_i}$
from Quotient Mapping is Linear Transformation.
That is: 
:$\map q x \in \span \set {\map q {x_1}, \ldots, \map q {x_n} }$ 
Since $x \in X$ was arbitrary, we deduce that $X/\ker T$ is generated by $\set {\map q {x_1}, \map q {x_2}, \ldots, \map q {x_n} }$.
So $X/\ker T$ is finite dimensional.
Category:Finite Rank Operators
Category:Quotient Vector Spaces