<The previous article in this series | The table of contents of this series |
description/proof of that for group and subgroup, quotient set by being in same coset is group only if subgroup is normal subgroup
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group and any subgroup, the quotient set by being in same coset is a group with respect to the canonical multiplication and inversion only if the subgroup is a normal subgroup.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(\sim_l\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in g' G)\}\), \(\in \{\text{ the equivalence relations }\}\)
\(\sim_r\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in G g')\}\), \(\in \{\text{ the equivalence relations }\}\)
\(G' / \sim_l\): \(= \text{ the quotient set }\)
\(G' / \sim_r\): \(= \text{ the quotient set }\)
//
Statements:
(
\(G' / \sim_l \text{ with the canonical multiplication and inversion } \in \{\text{ the groups }\}\)
\(\implies\)
\(G \in \{\text{ the normal subgroups }\}\)
)
\(\land\)
(
\(G' / \sim_r \text{ with the canonical multiplication and inversion } \in \{\text{ the groups }\}\)
\(\implies\)
\(G \in \{\text{ the normal subgroups }\}\)
)
//
2: Note
\(G' / \sim_l\) and \(G' / \sim_r\) are guaranteed to be well-defined as quotient sets, by the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
If \(G\) is a normal subgroup, \(G' / \sim_l = G' / \sim_r\) is a quotient group, by the proposition that with respect to any normal subgroup, the set of the cosets forms a group with the canonical multiplication and inversion; this proposition is claiming that \(G\) needs to be a normal subgroup in order for \(G' / \sim_l\) or \(G' / \sim_r\) to be a group with respect to the canonical multiplication and inversion.
In fact, if \(G\) is not any normal subgroup, the canonical multiplication and inversion are not well-defined.
3: Proof
Whole Strategy: Step 1: for \(G' / \sim_l\), suppose that the canonical multiplication is well-defined, let \(g \in G\) and \(g' \in G'\) be any, and see that \(g' g {g'}^{-1} \in G\); Step 2: for \(G' / \sim_r\), suppose that the canonical multiplication is well-defined, let \(g \in G\) and \(g' \in G'\) be any, and see that \(g' g {g'}^{-1} \in G\).
Step 1:
Let us suppose that for \(G' / \sim_l\), the canonical multiplication is well-defined.
"the canonical multiplication" means that \([g'_1] [g'_2] = [g'_1 g'_2]\).
Let \(g \in G\) and \(g' \in G'\) be any.
Let \(g'_1 \in G'\) be any.
Let \(g_1 \in G\) be any.
\([g'_1 g] = [g'_1]\), because \(g'_1, g'_1 g \in g'_1 G\).
\([{g'}^{-1} g_1] = [{g'}^{-1}]\), because \({g'}^{-1}, {g'}^{-1} g_1 \in {g'}^{-1} G\).
\([g'_1 g] [{g'}^{-1} g_1] = [g'_1] [{g'}^{-1}]\), which implies that \([g'_1 g {g'}^{-1} g_1] = [g'_1 {g'}^{-1}]\).
That means that \(g'_1 g {g'}^{-1} g_1 \in g'_1 {g'}^{-1} G\), which means that \(g'_1 g {g'}^{-1} g_1 = g'_1 {g'}^{-1} g_2\) for a \(g_2 \in G\).
So, \(g {g'}^{-1} g_1 = {g'}^{-1} g_2\), so, \(g' g {g'}^{-1} = g_2 {g_1}^{-1} \in G\).
That means that for each \(g' \in G'\), \(g' G {g'}^{-1} \subseteq G\).
By the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it, \(G\) is a normal subgroup.
Step 2:
Let us suppose that for \(G' / \sim_r\), the canonical multiplication is well-defined.
"the canonical multiplication" means that \([g'_1] [g'_2] = [g'_1 g'_2]\).
The logic is parallel to that for \(G' / \sim_l\).
Let \(g \in G\) and \(g' \in G'\) be any.
Let \(g'_1 \in G'\) be any.
Let \(g_1 \in G\) be any.
\([g g'_1] = [g'_1]\), because \(g'_1, g g'_1 \in G g'_1\).
\([g_1 g'] = [g']\), because \(g', g_1 g' \in G g'\).
\([g_1 g'] [g g'_1] = [g'] [g'_1]\), which implies that \([g_1 g' g g'_1] = [g' g'_1]\).
That means that \(g_1 g' g g'_1 \in G g' g'_1\), which means that \(g_1 g' g g'_1 = g_2 g' g'_1\) for a \(g_2 \in G\).
So, \(g_1 g' g = g_2 g'\), so, \(g' g {g'}^{-1} = {g_1}^{-1} g_2 \in G\).
That means that for each \(g' \in G'\), \(g' G {g'}^{-1} \subseteq G\).
By the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it, \(G\) is a normal subgroup.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for group and subgroup, being in same coset is equivalence relation
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(\sim_l\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in g' G)\}\)
\(\sim_r\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in G g')\}\)
//
Statements:
\(\sim_l \in \{\text{ the equivalence relations }\}\)
\(\land\)
\(\sim_r \in \{\text{ the equivalence relations }\}\)
//
2: Note
\(G\) does not need to be any normal subgroup.
So, the quotient sets, \(G' / \sim_l\) and \(G' / \sim_r\), are well-defined for any subgroup, \(G\), although \(G' / \sim_l\) and \(G' / \sim_r\) are not any groups if \(G\) is not any normal subgroup: refer to the proposition that for any group and any subgroup, the quotient set by being in same coset is a group only if the subgroup is a normal subgroup.
3: Proof
Whole Strategy: Step 1: see that \(\sim_l\) satisfies the conditions to be an equivalence relation; Step 2: see that \(\sim_r\) satisfies the conditions to be an equivalence relation.
Step 1:
Let us see that \(\sim_l\) satisfies the conditions to be an equivalence relation.
1) \(\forall g' \in G' (g' \sim_l g')\): reflexivity: \(g', g' \in g' G\).
2) \(\forall g'_1, g'_2 \in G' (g'_1 \sim_l g'_2 \implies g'_2 \sim_l g'_1)\): symmetry: if \(g'_1 \sim_l g'_2\), there is a \(g' \in G'\) such that \(g'_1, g'_2 \in g' G\), then, \(g'_2, g'_1 \in g' G\), so, \(g'_2 \sim_l g'_1\).
3) \(\forall g'_1, g'_2, g'_3 \in G' ((g'_1 \sim_l g'_2 \land g'_2 \sim_l g'_3) \implies g'_1 \sim g'_3)\): transitivity: if \(g'_1 \sim_l g'_2 \land g'_2 \sim_l g'_3\), there are a \(g'_4 \in G'\) such that \(g'_1, g'_2 \in g'_4 G\) and a \(g'_5 \in G'\) such that \(g'_2, g'_3 \in g'_5 G\), but as \(g'_2 \in g'_4 G, g'_5 G\), \(g'_2 = g'_4 g_1 = g'_5 g_2\) for some \(g_1, g_2 \in G\), so, \(g'_5 = g'_4 g_1 {g_2}^{-1}\), and as \(g'_3 = g'_5 g_3\) for a \(g_3 \in G\), \(g'_3 = g'_4 g_1 {g_2}^{-1} g_3 \in g'_4 G\), so, \(g'_1, g'_3 \in g'_4 G\).
Step 2:
Let us see that \(\sim_r\) satisfies the conditions to be an equivalence relation.
1) \(\forall g' \in G' (g' \sim_r g')\): reflexivity: \(g', g' \in G g'\).
2) \(\forall g'_1, g'_2 \in G' (g'_1 \sim_r g'_2 \implies g'_2 \sim_r g'_1)\): symmetry: if \(g'_1 \sim_r g'_2\), there is a \(g' \in G'\) such that \(g'_1, g'_2 \in G g'\), then, \(g'_2, g'_1 \in G g'\), so, \(g'_2 \sim_r g'_1\).
3) \(\forall g'_1, g'_2, g'_3 \in G' ((g'_1 \sim_r g'_2 \land g'_2 \sim_r g'_3) \implies g'_1 \sim g'_3)\): transitivity: if \(g'_1 \sim_r g'_2 \land g'_2 \sim_r g'_3\), there are a \(g'_4 \in G'\) such that \(g'_1, g'_2 \in G g'_4\) and a \(g'_5 \in G'\) such that \(g'_2, g'_3 \in G g'_5\), but as \(g'_2 \in G g'_4, G g'_5\), \(g'_2 = g_1 g'_4 = g_2 g'_5\) for some \(g_1, g_2 \in G\), so, \(g'_5 = {g_2}^{-1} g_1 g'_4\), and as \(g'_3 = g_3 g'_5\) for a \(g_3 \in G\), \(g'_3 = g_3 {g_2}^{-1} g_1 g'_4 \in G g'_4\), so, \(g'_1, g'_3 \in G g'_4\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for group, normal subgroup is kernel of canonical group homomorphism onto quotient group of group by normal subgroup
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group, any normal subgroup is the kernel of the canonical group homomorphism onto the quotient group of the group by the normal subgroup.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(G' / G\): \(= \text{ the quotient group }\)
\(f\): \(G' \to G' / G, g' \mapsto [g']\)
//
Statements:
\(f \in \{\text{ the group homomorphisms }\}\)
\(\land\)
\(G = \text{ the kernel of } f\)
//
2: Note
By the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain, the kernel of any group homomorphism is a normal subgroup of the domain; by this proposition, any normal subgroup of any group is the kernel of a group homomorphism.
So, the normal subgroup are the kernels of the group homomorphisms.
3: Proof
Whole Strategy: Step 1: see that \(f\) is a group homomorphism; Step 2: see that \(G\) is the kernel of \(f\).
Step 1:
Let us see that \(f\) is a group homomorphism.
\(f (1) = [1]\), which is the identity in \(G' / G\).
For each \(g'_1, g'_2 \in G'\), \(f (g'_1 g'_2) = [g'_1 g'_2] = [g'_1] [g'_2] = f (g'_1) f (g'_2)\).
For each \(g' \in G'\), \(f (g'^{-1}) = [g'^{-1}] = [g']^{-1} = f (g')^{-1}\).
So, \(f\) is a group homomorphism.
Step 2:
Let us see that \(G\) is the kernel of \(f\).
\(f (g') = [g'] = [1]\) means that \(g' G = 1 G\), which means that \(g' g = 1\) for a \(g \in G\), which means that \(g' = g^{-1} \in G\).
On the other hand, for each \(g \in G\), \(f (g) = [g] = [1]\), because \(g G = 1 G\), by the proposition that for any group, the multiplication map with any fixed element from left or right is a bijection.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Lie algebra, ideal is kernel of canonical Lie algebra homomorphism onto quotient Lie algebra of Lie algebra by ideal
Topics
About:
Lie algebra
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any Lie algebra, any ideal is the kernel of the canonical Lie algebra homomorphism onto the quotient Lie algebra of the Lie algebra by the ideal.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(V'\): \(\in \{\text{ the Lie algebras }\}\)
\(V\): \(\in \{\text{ the ideals of } V'\}\)
\(V' / V\): \(= \text{ the quotient Lie algebra }\)
\(f\): \(V' \to V' / V, v' \mapsto [v']\)
//
Statements:
\(f \in \{\text{ the Lie algebra homomorphisms }\}\)
\(\land\)
\(V = \text{ the kernel of } f\)
//
2: Note
By the proposition that the kernel of any Lie algebra homomorphism is an ideal of the domain, the kernel of any Lie algebra homomorphism is an ideal of the domain; by this proposition, any ideal of any Lie algebra is the kernel of a Lie algebra homomorphism.
So, the ideals are the kernels of the Lie algebra homomorphisms.
3: Proof
Whole Strategy: Step 1: see that \(f\) is a Lie algebra homomorphism; Step 2: see that \(V\) is the kernel of \(f\).
Step 1:
Let us see that \(f\) is a Lie algebra homomorphism.
Let \(v'_1, v'_2 \in V'\) and \(r_1, r_2 \in F\) be any.
\(f (r_1 v'_1 + r_2 v'_2) = [r_1 v'_1 + r_2 v'_2] = r_1 [v'_1] + r_2 [v'_2] = r_1 f (v'_1) + r_2 f (v'_2)\).
\(f ([v'_1, v'_2]) = [[v'_1, v'_2]] = [[v'_1], [v'_2]] = [f (v'_1), f (v'_2)]\).
So, \(f\) is a Lie algebra homomorphism.
Step 2:
Let us see that \(V\) is the kernel of \(f\).
\(f (v') = [0]\) means that \([v'] = [0]\), which means that \(v' \in V\).
On the other hand, for each \(v \in V\), \(f (v) = [v] = [0]\), so, \(v \in Ker (f)\).
So, \(V = Ker (f)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of quotient Lie algebra of Lie algebra by ideal of Lie algebra
Topics
About:
Lie algebra
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of quotient Lie algebra of Lie algebra by ideal of Lie algebra.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( V'\): \(\in \{\text{ the Lie algebras }\}\)
\( V\): \(\in \{\text{ the ideals of } V'\}\)
\(*V' / V\): \(= \text{ the quotient vectors space }\), with \([\bullet, \bullet]: V' / V \times V' / V \to V' / V\), \(\in \{\text{ the Lie algebras }\}\)
//
Conditions:
\(\forall [v'_1], [v'_2] \in V' / V ([[v'_1], [v'_2]] = [[v'_1, v'_2]])\)
//
2: Note
"\([[v'_1], [v'_2]] = [[v'_1, v'_2]]\)" may look confusing, but in "\([[v'_1], [v'_2]]\)", the inner brackets denote the equivalence classes in \(V' / V\) and the outer bracket denotes the bracket operation on \(V' / V\), while in "\([[v'_1, v'_2]]\)", the inner bracket denotes the bracket operation on \(V'\) and the outer bracket denotes the class in \(V' / V\): there should not be any other valid interpretation.
Let us see that the bracket on \(V' / V\) is indeed well-defined.
Let \([v'_1] = [v''_1]\) and \([v'_2] = [v''_2]\).
\(v''_1 = v'_1 + v_1\) and \(v''_2 = v'_2 + v_2\) for some \(v_1, v_2 \in V\).
\([[v''_1, v''_2]] = [[v'_1 + v_1, v'_2 + v_2]] = [[v'_1, v'_2] + [v'_1, v_2] + [v_1, v'_2] + [v_1, v_2]] = [[v'_1, v'_2] + [v'_1, v_2] - [v'_2, v_1] + [v_1, v_2]]\), but \([v'_1, v_2], [v'_2, v_1], [v_1, v_2] \in V\) because \(V\) is an ideal, so, \([v'_1, v_2] - [v'_2, v_1] + [v_1, v_2] \in V\), so, \( = [[v'_1, v'_2]]\).
Let us see that the bracket on \(V' / V\) satisfies the conditions for \(V' / V\) to be a Lie algebra.
Let \([v'_1], [v'_2], [v'_3] \in V' / V\) and \(r_1, r_2 \in F\) be any, where \(F\) is the field over which \(V'\) is a vectors space.
1) \([r_1 [v'_1] + r_2 [v'_2], [v'_3]] = [[r_1 v'_1 + r_2 v'_2], [v'_3]] = [[r_1 v'_1 + r_2 v'_2, v'_3]] = [r_1 [v'_1, v'_3] + r_2 [v'_2, v'_3]] = r_1 [[v'_1, v'_3]] + r_2 [[v'_2, v'_3]] = r_1 [[v'_1], [v'_3]] + r_2 [[v'_2], [v'_3]]\); \([[v'_3], r_1 [v'_1] + r_2 [v'_2]] = [[v'_3], [r_1 v'_1 + r_2 v'_2]] = [[v'_3, r_1 v'_1 + r_2 v'_2]] = [r_1 [v'_3, v'_1] + r_2 [v'_3, v'_2]] = r_1 [[v'_3, v'_1]] + r_2 [[v'_3, v'_2]] = r_1 [[v'_3], [v'_1]] + r_2 [[v'_3], [v'_2]]\).
2) \([[v'_2], [v'_1]] = [[v'_2, v'_1]] = [- [v'_1, v'_2]] = - [[v'_1, v'_2]] = - [[v'_1], [v'_2]]\).
3) \(\sum_{cyclic} [[v'_1], [[v'_2], [v'_3]]] = [[v'_1], [[v'_2], [v'_3]]] + [[v'_2], [[v'_3], [v'_1]]] + [[v'_3], [[v'_1], [v'_2]]] = [[v'_1], [[v'_2, v'_3]]] + [[v'_2], [[v'_3, v'_1]]] + [[v'_3], [[v'_1, v'_2]]] = [[v'_1, [v'_2, v'_3]]] + [[v'_2, [v'_3, v'_1]]] + [[v'_3, [v'_1, v'_2]]] = [[v'_1, [v'_2, v'_3]] + [v'_2, [v'_3, v'_1]] + [v'_3, [v'_1, v'_2]]] = [\sum_{cyclic} [v'_1, [v'_2, v'_3]]] = [0]\).
So, \(V' / V\) is s Lie algebra.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that kernel of Lie algebra homomorphism is ideal of domain
Topics
About:
Lie algebra
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the kernel of any Lie algebra homomorphism is an ideal of the domain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ Lie algebras }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ Lie algebras }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the Lie algebra homomorphisms }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
//
Statements:
\(Ker (f) \in \{\text{ the ideals of } V_1\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is a vectors subspace of \(V_1\); Step 2: see that \(Ker (f)\) satisfies the conditions to be an ideal.
Step 1:
\(Ker (f)\) is a vectors subspace of \(V_1\), by the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
Step 2:
Let \(v \in V_1\) and \(k \in Ker (f)\) be any.
\(f ([v, k]) = [f (v), f (k)]\), because \(f\) is a Lie algebra homomorphism, \(= [f (v), 0] = 0\), which means that \([v, k] \in Ker (f)\).
So, \(Ker (f)\) is an ideal of \(V_1\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of ideal of Lie algebra
Topics
About:
Lie algebra
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of ideal of Lie algebra.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( V'\): \(\in \{\text{ the Lie algebras }\}\)
\(*V\): \(\in \{\text{ the vectors subspaces of } V'\}\)
//
Conditions:
\(\forall v' \in V', \forall v \in V ([v', v] \in V)\)
//
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Hilbert space and its normed covectors (dual) space, there is canonical bijective complex-conjugate-linear isometry from Hilbert space onto normed covectors space
Topics
About:
Hilbert space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any Hilbert space and its normed covectors (dual) space, there is the canonical bijective complex-conjugate-linear isometry from the Hilbert space onto the normed covectors space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
\(V^*\): \(= \text{ the normed covectors space of } V\)
\(f\): \(: V \to V^*, v \mapsto \langle \bullet, v \rangle\)
//
Statements:
\(f \in \{\text{ the bijective complex-conjugate-linear isometries }\}\)
//
2: Note
The part of this proposition that each element of \(V^*\) is \(\langle \bullet, v \rangle\) for the unique \(v \in V\) is prevalently called "the Riesz representation theorem".
But this proposition is more than that: the Riesz representation theorem itself does not say that for each \(v \in V\), \(\langle \bullet, v \rangle \in V^*\), or that \(f\) is injective, or that \(f\) is complex-conjugate-linear, or that \(f\) is a 'normed vectors space' isometry.
3: Proof
Whole Strategy: Step 1: see that \(\langle \bullet, v \rangle \in V^*\); Step 2: see that \(f\) is bijective; Step 3: see that \(f\) is complex-conjugate-linear; Step 4: see that \(f\) is a 'normed vectors space' isometry.
Step 1:
Let us see that \(g: V \to F, v' \mapsto \langle v', v \rangle \in V^*\).
For each \(v_1, v_2 \in V\) and \(r_1, r_2 \in F\), \(g (r_1 v_1 + r_2 v_2) = \langle r_1 v_1 + r_2 v_2, v \rangle = r_1 \langle v_1, v \rangle + r_2 \langle v_2, v \rangle = r_1 g (v_1) + r_2 g (v_2)\), so, \(g\) is linear.
\(sup_{v' \in V \setminus \{0\}} \vert g (v') \vert / \Vert v' \Vert = \vert \langle v', v \rangle \vert / \sqrt{\langle v', v' \rangle} \le \sqrt{\langle v', v' \rangle} \sqrt{\langle v, v \rangle} / \sqrt{\langle v', v' \rangle}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, \(= \sqrt{\langle v, v \rangle} \lt \infty\), so, \(g\) is bounded.
So, \(g \in V^*\).
So, \(f\) is well-defined.
Step 2:
Let us see that \(f\) is injective.
Let \(v_1, v_2 \in V\) be any such that \(v_1 \neq v_2\).
Let us suppose that \(f (v_1) = f (v_2)\).
Then, \(\langle \bullet, v_1 \rangle = \langle \bullet, v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 \rangle = \langle v_1 - v_2, v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 \rangle - \langle v_1 - v_2, v_2 \rangle = 0\), but the left hand side was \(\langle v_1 - v_2, v_1 - v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 - v_2 \rangle = 0\), which would imply that \(v_1 - v_2 = 0\), so, \(v_1 = v_2\), a contradiction.
So, \(f (v_1) \neq f (v_2)\).
So, \(f\) is injective.
Let us see that \(f\) is surjective.
Let \(g \in V^*\) be any.
When \(g = 0\), \(g = \langle \bullet, 0 \rangle\).
Let us suppose that \(g \neq 0\) hereafter.
Let \(V^` := g^{-1} (0)\).
\(V^`\) is a vectors subspace of \(V\), by the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
\(V^`\) is a closed vectors subspace of \(V\), because \(\{0\}\) is a closed subset of \(F\), by the proposition that any linear map between any vectors metric spaces induced by any norms is continuous if and only if it is bounded and the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
Let \({V^`}^{\perp}\) be the orthogonal complement of \(V^`\).
\(V\) is a vectors space as direct sum of \(V^`\) and \({V^`}^{\perp}\), by the proposition that for any Hilbert space and its any closed vectors subspace, the space is the vectors space as direct sum of the subspace and its orthogonal complement.
As \(g \neq 0\), \(\{0\} \subset {V^`}^{\perp}\), because if \({V^`}^{\perp} = \{0\}\), for each \(v \in V\), \(v = v^` + 0\) where \(v^` \in V^`\), because \(V\) was a vectors space as direct sum of \(V^`\) and \({V^`}^{\perp}\), so, \(v = v^` \in V^`\), so, \(V^` = V\), which would mean that \(g = 0\), a contradiction.
So, there is a \(w \in {V^`}^{\perp}\) such that \(w \neq 0\), but let us take \(\Vert w \Vert = 1\), which is possible because \({V^`}^{\perp}\) is a vectors subspace: take \(w / \Vert w \Vert \in {V^`}^{\perp}\) instead.
For each \(v' \in V\), \(g (v') w - g (w) v' \in V^`\), because \(g (g (v') w - g (w) v') = g (v') g (w) - g (w) g (v') = 0\).
So, \(0 = \langle g (v') w - g (w) v', w \rangle = \langle g (v') w, w \rangle - \langle g (w) v', w \rangle = g (v') \langle w, w \rangle - g (w) \langle v', w \rangle = g (v') - \langle v', \overline{g (w)} w \rangle\), so, \(g (v') = \langle v', \overline{g (w)} w \rangle\).
That means that \(g (v') = \langle v', v \rangle\) where \(v := \overline{g (w)} w\), which means that \(g = \langle \bullet, v \rangle\).
So, \(f\) is surjective.
Step 3:
Let us see that \(f\) is complex-conjugate-linear.
Let \(v_1, v_2 \in V\) and \(r_1, r_2 \in F\) be any.
\(f (r_1 v_1 + r_2 v_2) = \langle \bullet, r_1 v_1 + r_2 v_2 \rangle = \overline{r_1} \langle \bullet, v_1 \rangle + \overline{r_2} \langle \bullet, v_2 \rangle = \overline{r_1} f (v_1) + \overline{r_2} f (v_2)\).
So, \(f\) is complex-conjugate-linear.
Step 4:
Let us see that \(f\) is a 'normed vectors space' isometry.
For each \(v \in V\), \(\Vert f (v) \Vert = sup_{v' \in V \setminus \{0\}} \vert f (v) (v') \vert / \Vert v' \Vert = sup_{v' \in V \setminus \{0\}} \vert \langle v', v \rangle \vert / \Vert v' \Vert \le sup_{v' \in V \setminus \{0\}} \sqrt{\vert \langle v', v' \rangle \vert} \sqrt{\vert \langle v, v \rangle \vert} / \sqrt{\vert \langle v', v' \rangle \vert}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, \(= sup_{v' \in V \setminus \{0\}} \sqrt{\vert \langle v, v \rangle \vert} = \sqrt{\vert \langle v, v \rangle \vert} = \Vert v \Vert\), and \(\vert f (v) (v') \vert / \Vert v' \Vert = \Vert v \Vert\) is in fact realized by \(v' = v\), because \(\vert f (v) (v) \vert / \Vert v \Vert = \vert \langle v, v \rangle \vert / \Vert v \Vert = \Vert v \Vert^2 / \Vert v \Vert = \Vert v \Vert\), so, \(\Vert f (v) \Vert = \Vert v \Vert\).
So, \(f\) is a 'normed vectors space' isometry.
References
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description/proof of that kernel of linear map between vectors spaces is vectors subspace of domain
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors fields }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors fields }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
//
Statements:
\(Ker (f) \in \{\text{ the vectors subspaces of } V_1\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is closed under linear combination; Step 2: conclude the proposition.
Step 1:
Let us see that \(Ker (f)\) is closed under linear combination.
Let \(v, v' \in Ker (f)\) and \(r, r' \in F\) be any.
\(f (r v + r' v') = r f (v) + r' f (v') = r 0 + r' 0 = 0\).
So, \(r v + r' v' \in Ker (f)\).
\(0 \in Ker (f)\).
Step 2:
By the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination, \(Ker (f)\) is a vectors subspace of \(V_1\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that kernel of linear map is submodule of domain
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the kernel of any linear map is a submodule of the domain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
//
Statements:
\(Ker (f) \in \{\text{ the submodules of } M_1\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is closed under linear combination; Step 2: conclude the proposition.
Step 1:
Let us see that \(Ker (f)\) is closed under linear combination.
Let \(m, m' \in Ker (f)\) and \(r, r' \in R\) be any.
\(f (r m + r' m') = r f (m) + r' f (m') = r 0 + r' 0 = 0\).
So, \(r m + r' m' \in Ker (f)\).
\(0 \in Ker (f)\).
Step 2:
By the proposition that for any module, any nonempty subset of the module is a submodule if and only if the subset is closed under linear combination, \(Ker (f)\) is a submodule of \(M_1\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of kernel of linear map
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of kernel of linear map.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( R\): \(\in \{\text{ the rings }\}\)
\( M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\( M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\( f\): \(: M_1 \to M_2\), \(\text{ the linear maps }\)
\(*Ker (f)\): \(= f^{-1} (0) \subseteq M_1\)
//
Conditions:
//
References
<The previous article in this series | The table of contents of this series | The next article in this series>
