nyuuzyou/wikis huggingface dataset:
Kernel is Trivial iff Monomorphism/Group

== Theorem ==
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then $\phi$ is a group monomorphism $\map \ker \phi$ is trivial.
== Proof ==
=== Necessary Condition ===
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group monomorphism.
By Homomorphism to Group Preserves Identity, $e_S \in \map \ker \phi$.
If $\map \ker \phi$ contained another element $s \ne e_S$, then $\map \phi s = \map \phi {e_S} = e_T$ and $\phi$ would not be injective, thus not be a group monomorphism.
So $\map \ker \phi$ can contain only one element, and that must be $e_S$, which is therefore the trivial subgroup of $S$.
=== Sufficient Condition ===
Now suppose $\map \ker \phi = \set {e_S}$.
Then, for any $x, y \in S$:
Thus $\phi$ is injective, and therefore a group monomorphism. 
== Sources ==
* : $\S 7.4$. Kernel and image: Example $139$
* : $\S 1.10$: Theorem $22$
* : Chapter $\text{II}$: Groups: Quotient Groups
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 65$
* : $\S 1.2$
* : $\S 47.8$ Homomorphisms and their elementary properties
* : Chapter $8$: The Homomorphism Theorem: Exercise $1$
* : Chapter $13$: Corollary $13.18$
Category:Kernel is Trivial iff Monomorphism
Category:Group Monomorphisms
Kernel is Trivial iff Monomorphism/Ring

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then $\phi$ is a ring monomorphism $\map \ker \phi = 0_{R_1}$.
== Proof ==
The proof for the ring monomorphism follows directly from:
:Ring Homomorphism of Addition is Group Homomorphism
and:
:Kernel is Trivial iff Group Monomorphism for the group monomorphism.
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $45$
* : $\S 2.2$: Homomorphisms: Lemma $2.6 \ \text{(i)}$
* : $\S 57.2$ Ring homomorphisms
Category:Kernel is Trivial iff Monomorphism
Category:Ring Monomorphisms
Kernel of Adjoint of Densely-Defined Linear Operator is Orthocomplement of Image

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
Then we have:
:$\map \ker {A^\ast} = \paren {\Img T}^\bot$
where $\bot$ denotes orthocomplementation.
== Proof ==
Let $y \in \HH$.
Define $f_y : \map D T \to \C$ by:
:$\map {f_y} x = \innerprod {A x} y$ for each $x \in \map D T$.
We have $y \in \map D {T^\ast}$ $f_y$ is a bounded linear functional.
We have $y \in \paren {\Img T}^\bot$ :
:$\map {f_y} x = \innerprod {A x} y = 0$ for all $x \in \map D T$.
Hence if $y \in \paren {\Img T}^\bot$, we have $y \in \map D {T^\ast}$.
Conversely if $\map {f_y} x = 0$ for all $x \in \map D T$ we have $y \in \map D {T^\ast}$ and $y \in \paren {\Img T}^\bot$.
Hence we have $y \in \paren {\Img T}^\bot$ :
:$0 = \innerprod {A x} y = \innerprod x {A^\ast y}$ for all $x \in \map D T$.
This is equivalent to:
:$A^\ast y \in \map D T^\bot$
From Linear Subspace Dense iff Zero Orthocomplement, we have $\map D T^\bot = \set { {\mathbf 0}_\HH}$.
Hence this is equivalent to $A^\ast y = {\mathbf 0}_\HH$.
This is equivalent to $y \in \map \ker {A^\ast}$.
Hence we have:
:$\map \ker {A^\ast} = \paren {\Img T}^\bot$
Category:Adjoints (Densely-Defined Linear Operators)
Category:Orthocomplements
Kernel of Bounded Linear Transformation is Annihilator of Image of Dual Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively. 
Let $T : X \to Y$ be a Bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Then we have: 
:$\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$
where ${}^\bot$ denotes the annihilator operation on subspaces of $X^\ast$. 
== Proof ==
Let $x \in \map \ker T$. 
Let $\phi \in T^\ast \sqbrk {Y^\ast}$.
Then from the definition of the dual operator, there exists $f \in Y^\ast$ such that $f \circ T = \phi$.
Then we have: 
Since this holds for all $\phi \in T^\ast \sqbrk {Y^\ast}$, we obtain $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.
So we have:
:$\map \ker T \subseteq {}^\bot T^\ast \sqbrk {Y^\ast}$
Now let $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.
Then for all $\phi \in T^\ast \sqbrk {Y^\ast}$ we have $\map \phi x = 0$. 
That is, for all $f \in Y^\ast$, we have: 
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$
from the definition of the dual operator.
From Normed Dual Space Separates Points, we obtain $T x = {\mathbf 0}_Y$ and hence $x \in \map \ker T$. 
So we obtain:
:${}^\bot T^\ast \sqbrk {Y^\ast} \subseteq \map \ker T$
Hence we can conclude:
:$\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$
== Sources ==
* : Exercise $2.39 \text { (i)}$
Category:Annihilators of Subspaces of Normed Dual Spaces
Category:Dual Operators
Kernel of Bounded Linear Transformation is Closed Linear Subspace

== Theorem ==
Let $V, W$ be normed vector spaces.
Let $f: V \to W$ be a bounded linear transformation.
Then $\ker f$, the kernel of $f$, is a closed linear subspace of $V$.
== Proof ==
By Kernel of Linear Transformation is Linear Subspace, $\ker f$ is a subspace of $V$.
By Continuity of Linear Transformations, $f$ is continuous.
Since $\ker f = f^{-1} \sqbrk{ \set{ \mathbf 0_W } }$, it follows from Continuity Defined from Closed Sets that $\ker f$ is closed.
Hence the result by definition of closed linear subspace.
Category:Bounded Linear Transformations
Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital commutative Banach algebra over $\C$. 
Let $\phi : A \to \C$ be a character on $A$. 
Then $\ker \phi$ is a maximal ideal of $A$.
== Proof 1 ==
== Proof 2 ==
Category:Characters (Banach Algebras)
Category:Commutative Banach Algebras
Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal/Proof 1

== Theorem ==
== Proof ==
From Kernel of Ring Homomorphism is Ideal, $\ker \phi$ is a ring ideal of $A$. 
From the First Ring Isomorphism Theorem, we have:
:$\phi \sqbrk A$ and $\dfrac A {\ker \phi}$ are isomorphic as rings.
From Character on Banach Algebra is Surjective, we have that $\phi \sqbrk A = \C$.
Hence:
:$\dfrac A {\ker \phi} \cong \C$
That is:
:$\dfrac A {\ker \phi}$ is a field.
From Maximal Ideal iff Quotient Ring is Field, we conclude that $\ker \phi$ is a maximal ideal of $A$ in the sense of a ring.
From Kernel of Linear Transformation is Linear Subspace, $\ker \phi$ is a maximal ideal in the sense of an algebra as well.
Category:Characters (Banach Algebras)
Category:Commutative Banach Algebras
Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal/Proof 2

== Theorem ==
== Proof ==
Let $I$ be an ideal of $A$ such that:
:$\ker \phi \subsetneq I$
We need to show $I = A$.
That is, we need to show:
:${\mathbf 1}_A \in I$
Let:
:$x \in I \setminus \ker \phi$
Then:
:$\map \phi x \ne 0$
Thus we can define:
:$\ds \tilde x := {\map \phi x}^{-1} x$
Then:
Thus:
:${\mathbf 1}_A = \underbrace{ {\mathbf 1}_A - \tilde x}_{\in \ker \phi} + \underbrace{\paren { {\map \phi x}^{-1} {\mathbf 1}_A} }_{\in A} \underbrace{x}_{\in I} \in I$
Category:Characters (Banach Algebras)
Category:Commutative Banach Algebras
Kernel of Closed Linear Transformation is Closed

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces.
Let $T : \map D T \to Y$ be a closed linear transformation.
Let $\ker T$ be the kernel of $T$.
Then $\ker T$ is closed. 
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\ker T$ converging to $x \in X$. 
We have $T x_n = {\mathbf 0}_Y$ for each $n \in \N$.
Hence from Constant Sequence in Normed Vector Space Converges, we have $T x_n \to {\mathbf 0}_Y$ as $n \to \infty$.
From Sequential Characterization of Closed Linear Transformation, we therefore have that $x \in \map D T$ and:
:$\ds T x = \lim_{n \mathop \to \infty} T x_n = {\mathbf 0}_Y$
Hence $x \in \ker T$. 
So every convergent sequence in $\ker T$ has its limit in $\ker T$.
Hence $\ker T$ is closed.
Category:Closed Linear Transformations
Kernel of Composition of Ring Homomorphisms

== Theorem ==
Let $f : A \to B$ and $g : B \to C$ be ring homomorphisms.
Let $g \circ f$ be their composition.
Then the kernel of $g \circ f$ is the preimage under $f$ of the kernel of $g$:
:$\map \ker {g \circ f} = f^{-1} \sqbrk {\ker g}$
== Proof ==
By definition, the kernel of a ring homomorphism is its kernel when considered as a group homomorphism between the additive groups.
The result follows from Kernel of Composition of Group Homomorphisms.
Category:Ring Homomorphisms
Kernel of First Projection on Group Direct Product is Isomorphic to Second Factor

== Theorem ==
Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups.
Let $\struct {G, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Let $\pr_1$ denote the first projection of $\struct {G, \circ}$:
:$\forall \tuple {x, y} \in G: \map \phi {x, y} = x$
Then the kernel of $\pr_1$ is isomorphic to $\struct {G_2, \circ_2}$.
== Proof ==
It is established in Projection on Group Direct Product is Epimorphism that $\pr_1$ is an epimorphism and so a homomorphism.
The kernel of $\pr_1$ is, by definition:
Let $H := \set {\tuple {e_1, y}: y \in G_2}$.
It is to be established that $\struct {H, \circ}$ is isomorphic to $\struct {G_2, \circ_2}$.
Let $\theta: H \to G_2$ be defined as:
:$\forall \tuple {e_1, y} \in H: \map \theta {e_1, y} = y$
We have that $\theta$ is the second projection on $H$, and so itself an epimorphism, and so surjective.
Then:
demonstrating that $\theta$ is an injection.
So $\theta$ is a homomorphism which is both surjective and injective.
Therefore $\theta$ is an isomorphism from $\map \ker {\pr_1}$ to $\struct {G_2, \circ_2}$.
Hence the result.
== Sources ==
* : $\text{II}$: Direct Products
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $5$
Category:Group Epimorphisms
Category:Group Direct Products
Category:Projection on Group Direct Product is Epimorphism
Kernel of Group Action is Normal Subgroup

== Theorem ==
Let $\struct{G, \cdot}$ be a group whose identity is $e$.
Let $X$ be a set.
Let $\phi: G \times X \to X$ be a group action such that:
:$\map \phi x = g * x$
Let $G_0$ denote the kernel of $\phi$.
Then $G_0$ is a normal subgroup of $G$.
== Proof 1==
Let $h \in G_0$.
 
From Stabilizer is Subgroup:
:$\Stab x \le G$
Thus $G_0$ is the intersection of subgroups.
By Intersection of Subgroups is Subgroup:
:$G_0 \le G$
Let $g \circ G_0$ denote the subset product of $g$ with $G_0$.
To prove normality it is sufficient to show:
:$\forall g \in G: g \circ G_0 \circ g^{-1} = G_0$
Let $h \in G_0, g \in G$ be arbitrary.
We have that:
:$h \in \Stab { g^{-1} * x}$
Then:
Therefore:
:$g \cdot h \cdot g^{-1} \in G_0$
so:
:$g \circ G_0 \circ g^{-1} \subseteq G_0$
Conversely suppose that:
:$h \in G_0$
Then by the above:
:$h' = g^{-1} \cdot h \cdot g \in G_0$
Therefore:
:$h = g \cdot h' \cdot g^{-1} \in g \circ G_0 \circ g^{-1}$
and so:
:$G_0 \subseteq g \circ G_0 \circ g^{-1}$
This concludes the proof.
== Proof 2==
We are given $\phi: G \times X \to X$ is a group action such that:
:$\map \phi x = g * x$
We are given $G_0$ is the kernel of $\phi$.
Let $\phi_g : X \to X$ be the mapping:
:$\map {\phi_g} x = \map \phi {g, x}$
for any $g \in G$
By Definition 2 of Kernel of Group Action, the kernel of the group action is the kernel of its permutation representation.
By definition, the permutation representation of $G$ associated to the group action is the group homomorphism $\rho : G \to \struct {\map \Gamma X, \circ}$ which sends $g$ to $\phi_g$.
By Correspondence Between Group Actions and Permutation Representations, we have that $\rho$ is the permutation representation associated to $\phi$
Let $\tilde \phi: G \to \map \Gamma X$ be the permutation representation associated to $\phi$, defined by:
:$\map {\tilde \phi} g := \phi_g$
By Group Action defines Permutation Representation $\map {\tilde \phi} g$ is a group homomorphism.
By Group Homomorphism Preserves Identity, $\map {\tilde \phi} e$ is the identity element for the kernel of the permutation representation of $G$ associated to the group action.
By Inner Automorphism Maps Subgroup to Itself iff Normal we have that:
:$\forall x \in G_0: \kappa_x \sqbrk {G_0} = G_0$
Hence the result.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 53 \beta$
Category:Group Actions
Kernel of Group Homomorphism Corresponds with Normal Subgroup of Domain

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then there exists $N \lhd G$, a normal subgroup of $G$ such that:
:$N = \map \ker \phi$
Conversely, let $N \lhd G$ be normal subgroup of $G$.
Then there exists $\phi: \struct {G, \circ} \to \struct {H, *}$, a group homomorphism, whose kernel $\map \ker \phi$ is such that:
:$\map \ker \phi = N$
== Proof ==
The first statement is Kernel is Normal Subgroup of Domain:
:The kernel of $\phi$ is a normal subgroup of the domain of $\phi$:
::$\map \ker \phi \lhd \Dom \phi$
The second statement is Quotient Group Epimorphism is Epimorphism:
:The mapping $\phi: G \to G / N$, defined as:
::$\phi: G \to G / N: \map \phi x = x N$
:is a group epimorphism, and its kernel is $N$.
== Sources ==
* : $\S 1.10$: Theorem $25$
* : Chapter $\text{II}$: Groups: Quotient Groups: Theorem $3$
* 
Category:Normal Subgroups
Category:Group Homomorphisms
Category:Kernels of Group Homomorphisms
Kernel of Group Homomorphism is Subgroup

== Theorem ==
The kernel of a group homomorphism is a subgroup of its domain:
:$\map \ker \phi \le \Dom \phi$
== Proof ==
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
From Homomorphism to Group Preserves Identity, $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$.
Therefore $\map \ker \phi \ne \O$.
Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y = e_H$.
Then:
So $x^{-1} \circ y \in \map \ker \phi$, and from the One-Step Subgroup Test, $\map \ker \phi \le \Dom \phi$.
== Also see ==
* Kernel is Normal Subgroup of Domain
== Sources ==
* : $\S 7.4$. Kernel and image
* : $\S 1.10$: Theorem $22$
* : Chapter $\text{II}$: Groups: Morphisms
* : $\S 1.2$
* : $\S 47.7$ Homomorphisms and their elementary properties
Category:Subgroups
Category:Kernels of Group Homomorphisms
Kernel of Group Homomorphism is not Empty

== Theorem ==
Let $G$ and $H$ be groups whose identity elements are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a homomorphism from $G$ to $H$.
Let $\map \ker \phi$ denote the kernel of $\phi$.
Then:
:$\map \ker \phi \ne \O$
where $\O$ denotes the empty set.
== Proof ==
From Identity is in Kernel of Group Homomorphism we have that:
:$e_G \in \map \ker \phi$
Hence the result.
== Proof ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 65$
Category:Kernels of Group Homomorphisms
Kernel of Homomorphism on Cyclic Group

== Theorem ==
Let $G = \gen g$ be a cyclic group with generator $g$.
Let $H$ be a group.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $\map \ker \phi$ denote the kernel of $\phi$.
Let $\Img G$ denote the homomorphic image of $G$ under $\phi$.
Then:
:$\map \ker \phi = \gen {g^m}$
where:
:$m = 0$ if $\Img \phi$ is an infinite cyclic group
:$m = \order {\Img \phi}$ if $\Img \phi$ is a finite cyclic group.
== Proof ==
From Kernel of Group Homomorphism is Subgroup and Subgroup of Cyclic Group is Cyclic:
:$\exists m \in \N: \map \ker \phi = \gen {g^m}$
From Homomorphic Image of Cyclic Group is Cyclic Group:
:$\Img \phi$ is a cyclic group generated by $\map \phi g$.
=== Case $1$: $\Img \phi$ is infinite ===
 $m \ne 0$.
Then $g^m$ is not the identity.
Thus:
However this contradicts $g^m \in \gen {g^m} = \map \ker \phi$.
Hence we must have $m = 0$.
=== Case $2$: $\Img \phi$ is finite ===
From Order of Cyclic Group equals Order of Generator, the order of $\map \phi g$ is $\order {\Img \phi}$.
Let $n \in \Z$.
By Division Theorem:
:$\exists q, r \in \Z: 0 \le r < \order {\Img \phi}: n = q \order {\Img \phi} + r$
We have:
From definition of order of group element:
:$0 < r < \order {\Img \phi} \implies \paren {\map \phi g}^r \ne e_H$
Hence:
and we see that $m = \order {\Img \phi}$ satisfies the above relation.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $8$
Category:Cyclic Groups
Category:Kernels of Group Homomorphisms
Kernel of Induced Homomorphism of Polynomial Forms

== Theorem ==
Let $R$ and $S$ be commutative rings with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $K = \ker \phi$.
Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.
Let $\bar \phi: R \sqbrk X \to S \sqbrk X$ be the induced morphism of polynomial rings.
Then the kernel of $\bar \phi$ is:
:$\ker \bar \phi = \set {a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X: \map \phi {a_i} = 0 \text{ for } i = 0, \ldots, n}$
Or, more concisely:
:$\ker \bar \phi = \paren {\ker \phi} \sqbrk X$
== Proof ==
Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Suppose first that $\map \phi {a_i} = 0$ for $i = 0, \ldots, n$.
We have by definition that:
:$\map {\bar \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n = 0$
That is to say, $\map P X \in \ker \bar \phi$.
Conversely, suppose that $\map P X \in \ker \bar \phi$.
That is, $\map {\bar \phi} {\map P X} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n$ is the null polynomial.
This by definition means that for $i = 0, \ldots, n$ we have $\map \phi {a_i} = 0$.
Hence, $\map P X \in \paren {\ker \phi} \sqbrk X$.
This concludes the proof.
Category:Polynomial Theory
Kernel of Inner Automorphism Group is Center

== Theorem ==
Let the mapping $\kappa: G \to \Inn G$ from a group $G$ to its inner automorphism group $\Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the inner automorphism of $G$ given by $a$.
Then $\kappa$ is a group epimorphism, and its kernel is the center of $G$:
:$\map \ker \kappa = \map Z G$
== Proof ==
Let $\kappa: G \to \Aut G$ be a mapping defined by $\map \kappa x = \kappa_x$.
It is clear that $\Img \kappa = \Inn G$.
It is also clear that $\kappa$ is a homomorphism:
Note that $\forall \kappa_x \in \Inn G: \exists x \in G: \map \kappa x = \kappa_x$.
Thus $\kappa: G \to \Inn G$ is a surjection and therefore an group epimorphism.
Now we investigate the kernel of $\kappa$:
So the kernel of $\kappa$ is the center of $G$.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $10$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.11 \ \text{(b)}$
* : Chapter $\text{II}$: Groups: Problem $\text{AA}$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$
* : Chapter $8$: The Homomorphism Theorem: Proposition $8.17$
Category:Inner Automorphisms
Category:Centers of Groups
Kernel of Linear Transformation between Finite-Dimensional Normed Vector Spaces is Closed

== Theorem ==
Let $m, n \in \N_{> 0}$ be natural numbers.
Let $A \in \R^{m \times n}$ be a matrix.
Let $\ker A = \set {\mathbf x \in \R^n : A \mathbf x = 0}$ be the kernel of $A$.
Then $\ker A$ is a closed subspace of $\R^n$.
== Proof ==
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
By Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, $T_A : \R^n \to \R^m$ is continuous. 
We have that Singleton in Normed Vector Space is Closed. 
Hence, $\set {\mathbf 0}$ is closed in $\R^n$.
By the corollary of Mapping is Continuous iff Inverse Images of Open Sets are Open, the inverse image of $\set {\mathbf 0}$ under $T_A$ is closed in $\R^n$.
However:
:$\map {T^{-1}_A} {\set {\mathbf 0}} = \set {\mathbf x \in \R^n : A \mathbf x = 0} = \ker A$
Hence, $\ker A$ is closed in $\R^n$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Linear Transformations
Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional

== Theorem ==
Let $V$ be a complex vector space.
Let $\map \LL {V, \C}$ be the space of all linear transformations from $V$ to complex numbers $\C$.
Let $\ell, L \in \map \LL {V, \C}$ be such that:
:$\ker \ell \subseteq \ker L$
where $\ker$ denotes the kernel.
Then:
:$\exists c \in \C : L = c \ell$
== Proof ==
Suppose $\ell = \mathbf 0$.
Then:
:$\ker \ell = V$
That is, the kernel of $\ell$ is the entire vector space $V$.
Moreover:
:$\ker \ell \subseteq \ker L \implies \ker L = V$
Therefore:
:$L = \mathbf 0$
and we can set $c = 0$ to have:
:$L = \mathbf 0 = 0 \cdot \ell$
Suppose $\ell \ne \mathbf 0$.
By Linear Transformation Maps Zero Vector to Zero Vector:
:$\exists v_0 \in V : v_0 \ne \mathbf 0 : \map \ell {v_0} \ne 0$
Let $v \in V$.
Let:
:$w = v - c_v v_0$
where $c_v \in \C$.
Then:
Suppose $w \in \ker \ell$.
Then:
:$w \in \ker L$
and it follows that:
:$\map \ell w = \map L w = 0$
Now we will seek for $w$ such that $\map \ell w = 0$.
Hence, if $\ds c_v = \frac {\map \ell v} {\map \ell {v_0} }$ then:
Therefore:
:$\ds c = \frac {\map L {v_0} }{\map \ell {v_0} }$
== Sources ==
* : Chapter $\S 6.2$: Derivatives in the distributional sense
Category:Linear Transformations
Kernel of Linear Transformation contains Zero Vector

== Corollary to Linear Transformation Maps Zero Vector to Zero Vector ==
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:
:$\mathbf 0 \in \map \ker T$
where $\map \ker T$ is the kernel of $T$.
== Proof ==
Follows from Linear Transformation Maps Zero Vector to Zero Vector and the definition of kernel.
Category:Linear Transformations
Category:Linear Transformation Maps Zero Vector to Zero Vector
Kernel of Linear Transformation is Linear Subspace

== Theorem ==
Let $V, W$ be normed vector spaces on a field $F$.
Let $T: V \to W$ be a linear transformation.
Then the kernel of $T$ is a linear subspace of $V$.
== Proof ==
By Kernel of Linear Transformation contains Zero Vector, $0_V \in \ker T$.
=== Closure under Addition ===
For $x, y \in \ker T$:
:$\map T {x + y} = \map T x + \map T y = 0_W + 0_W = 0_W$
so:
:$x + y \in \ker T$
so $\ker T$ is closed under addition.
=== Closure under Scalar Multiplication ===
For $k \in F, x \in \ker T$:
:$\map T {k x} = k \map T x = k 0_W = 0_W$
so:
:$k x \in \ker T$
So $\ker T$ is closed under scalar multiplication.
Hence the result by definition of linear subspace.
Category:Linear Transformations