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description/proof of that for Lie group homomorphism, if differential at identity Is \(0\), map is constant over connected component
Topics
About:
\(C^\infty\) manifold
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 group homomorphism, if the differential at the identity is \(0\), the map is constant over each connected component.
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:
\(G_1\): \(\in \{\text{ the Lie groups }\}\)
\(G_2\): \(\in \{\text{ the Lie groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(d f_1\): \(: T_1{G_1} \to T_1{G_2}\), \(= \text{ the differential at } 1\)
//
Statements:
\(d f_1 = 0\)
\(\implies\)
\(\forall T \in \{\text{ the connected components of } M_1\} (f \vert_T \in \{\text{ the constant maps }\})\)
//
2: Note
Especially, for the connected component that contains \(1\), \(T_1\), \(f \vert_T = 1\), because \(f \vert_T (1) = 1\).
3: Proof
Whole Strategy: Step 1: see that \(f\) is a constant rank map; Step 2: conclude the proposition.
Step 1:
Any Lie group homomorphism has a constant rank, by the proposition that for any 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank: the immediate corollary mentioned in the Note.
So, \(f\) has a constant rank, but as \(f\) has the rank, \(0\), at \(1\), \(f\) has the constant \(0\) rank.
Step 2:
Let \(T\) be any connected component of \(M_1\).
By the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds, if the map has the constant rank \(0\), the map is constant over each connected component, \(f \vert_T\) is constant.
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 2 Lie group homomorphisms between Lie groups, map as multiplication of 1st homomorphism and multiplicative inverse of 2nd homomorphism has constant rank
Topics
About:
\(C^\infty\) manifold
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 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank.
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:
\(G_1\): \(\in \{\text{ the Lie groups }\}\)
\(G_2\): \(\in \{\text{ the Lie groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(f'\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(h\): \(: G_1 \to G_2, g \mapsto f (g) (f' (g))^{-1}\)
//
Statements:
\(\forall g \in G_1 (Rank ({d h}_g) = Rank ({d h}_1))\)
//
2: Note
\(h\) is not any Lie group homomorphism in general, because \(h (g g') = f (g g') (f' (g g'))^{-1} = f (g) f (g') (f' (g) f' (g'))^{-1} = f (g) f (g') (f' (g'))^{-1} (f' (g))^{-1} = f (g) h (g') (f' (g))^{-1}\), which is not guaranteed to equal \(h (g) h (g') = f (g) (f' (g))^{-1} f (g') (f' (g'))^{-1}\), unless \(G_2\) is Abelian.
As an immediate corollary, \(f\) has a constant rank, because \(f'\) can be taken to be \(f' = 1\), which is a Lie group homomorphism, and \(h (g) = f (g) (f' (g))^{-1} = f (g) 1^{-1} = f (g) 1 = f (g)\), so, \(h = f\).
3: Proof
Whole Strategy: Step 1: see that \(h (g_0 g) = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h (g)\) and take \(h': G_1 \to G_2, g \mapsto h (g_0 g), = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h = h \circ L_{g_0}\); Step 2: see that \({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0)} \circ {d L_{f (g_0)}}_1 \circ {d h}_1 = {d h}_{g_0} \circ {d L_{g_0}}_1\); Step 3: conclude the proposition.
Step 1:
\(h\) is \(C^\infty\), because \(f\) and \(f'\) are \(C^\infty\), and the multiplicative inverse map and the multiplication map are \(C^\infty\): it is \(: G_1 \to G_1 \times G_1 \to G_2 \times G_2 \to G_2 \times G_2 \to G_2, g \mapsto (g, g) \mapsto (f (g), f' (g)) \mapsto (f (g), (f' (g))^{-1}) \mapsto f (g) (f' (g))^{-1}\).
For each \(g_0 \in G_j\), let \(L_{g_0}: G_j \to G_j, g \mapsto g_0 g\) and \(R_{g_0}: G_j \to G_j, g \mapsto g g_0\) be the left translation and the right translation by \(g_0\), which are some diffeomorphisms.
For each \(g_0, g \in G_1\), let us see that \(h (g_0 g) = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h (g)\).
\(h (g_0 g) = f (g_0 g) (f' (g_0 g))^{-1} = f (g_0) f (g) (f' (g_0) f' (g))^{-1} = f (g_0) f (g) (f' (g))^{-1} (f' (g_0))^{-1} = f (g_0) h (g) (f' (g_0))^{-1} = L_{f (g_0)} (h (g)) (f' (g_0))^{-1} = R_{(f' (g_0))^{-1}} (L_{f (g_0)} (h (g)))\).
So, let us take \(h': G_1 \to G_2, g \mapsto h (g_0 g), = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h = h \circ L_{g_0}\), which is \(C^\infty\) as a composition of \(C^\infty\) maps.
Step 2:
\({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0) h (1)} \circ {d L_{f (g_0)}}_{h (1)} \circ {d h}_1 = {d h}_{g_0 1} \circ {d L_{g_0}}_1\).
As \(h (1) = f (1) (f' (1))^{-1} = 1 1^{-1} = 1 1 = 1\), \({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0)} \circ {d L_{f (g_0)}}_1 \circ {d h}_1 = {d h}_{g_0} \circ {d L_{g_0}}_1\).
Step 3:
\({d R_{(f' (g_0))^{-1}}}_{f (g_0)}: T_{f (g_0)}G_2 \to T_{f (g_0) (f' (g_0))^{-1}}G_2\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(R_{(f' (g_0))^{-1}}\) is a diffeomorphism.
\({d L_{f (g_0)}}_1: T_1G_2 \to T_{f (g_0)}G_2\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(L_{f (g_0)}\) is a diffeomorphism.
\({d h}_1: T_1G_1 \to T_1G_2\).
\({d h}_{g_0}: T_{g_0}G_1 \to T_{h (g_0)}G_2\).
\({d L_{g_0}}_1: T_1G_1 \to T_{g_0}G_1\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(L_{g_0}\) is a diffeomorphism.
By the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map, \(Rank ({d h'}_1) = Rank ({d h}_1) = Rank ({d h}_{g_0})\).
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 \(C^\infty\) map between \(C^\infty\) manifolds, if map has constant rank \(0\), map is constant over connected component
Topics
About:
\(C^\infty\) manifold
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 \(C^\infty\) map between any \(C^\infty\) manifolds, if the map has the constant rank \(0\), the map is constant over each connected component.
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:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
//
Statements:
\(\forall m \in M_1 (Rank (f)_m = 0)\)
\(\implies\)
\(\forall T \in \{\text{ the connected components of } M_1\} (f \vert_T \in \{\text{ the constant maps }\})\)
//
2: Note
\(M_1\) and \(M_2\) are required to be without boundary for this proposition, because this proposition uses the rank theorem for \(C^\infty\) map between \(C^\infty\) manifolds, which requires the domain and the codomain without boundary.
As an immediate corollary, when \(M_1\) is connected, \(f\) is constant.
3: Proof
Whole Strategy: Step 1: see that around each \(m \in M_1\), there is an open neighborhood, \(U_m\), over which \(f\) is constant; Step 2: conclude the proposition.
Step 1:
Let \(m \in M_1\) be any.
As \(f\) has the constant rank, \(0\), by the rank theorem for \(C^\infty\) map between \(C^\infty\) manifolds, there are a chart around \(m\), \((U_m \subseteq M_1, \phi_m)\), and a chart around \(f (m)\), \((U_{f (m)} \subseteq M_2, \phi_{f (m)})\), such that \(\phi_{f (m)} \circ f \circ {\phi_m}^{-1} \vert_{\phi_m (U_m)}: \phi_m (U_m) \to \phi_{f (m)} (U_{f (m)})\) is \((x^1, ..., x^{d_1}) \mapsto (0, ..., 0)\).
That means that \(f\) is constant over \(U_m\).
Step 2:
Let \(T \subseteq M_1\) be any connected component of \(M_1\).
\(T\) is a connected topological subspace of \(M_1\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
\(f \vert_T: T \to M_2\) is a map that is anywhere locally constant over a connected topological space: around each \(t \in T\), there is an open neighborhood of \(t\) on \(M_1\), \(U_t \subseteq M_1\), over which \(f\) is constant by Step 1, and \(U_t \cap T \subseteq T\) is an open neighborhood of \(t\) on \(T\) over which \(f \vert_T\) is constant.
By the proposition that any map that is anywhere locally constant on any connected topological space is globally constant, \(f \vert_T\) is globally constant.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of rank of \(C^\infty\) map between \(C^\infty\) manifolds with boundary at point
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of rank of \(C^\infty\) map between \(C^\infty\) manifolds with boundary at point.
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:
\( M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary } \}\)
\( M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary } \}\)
\( f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\( m\): \(\in M_1\)
\( d f_m\): \(: T_mM_1 \to T_{f (m)}M_2\), \(= \text{ the differential of } f \text{ at } m\)
\(*Rank (f)_m\): \(= Rank (d f_m)\)
//
Conditions:
//
2: Note
In general, the rank of \(f\) changes point to point.
When the rank of \(f\) is the same at all the points of \(M_1\), \(f\) is called to have "constant rank".
When at each point of \(M_1\), there is a neighborhood of the point over which \(f\) has the same rank, \(f\) is called to have "locally constant rank".
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 linear map between vectors spaces, 'vectors spaces - linear morphisms' isomorphism onto domain of linear map, and 'vectors spaces - linear morphisms' isomorphism from superspace of codomain of linear map, composition of linear map after 1st isomorphism and before 2nd isomorphism has rank of linear map
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 for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_1\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(V_0\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_0\): \(: V_0 \to V_1\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
\(V'_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), such that \(V_2 \in \{\text{ the vectors subspaces of } V'_2\}\)
\(V_3\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_2\): \(: V'_2 \to V_3\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
Statements:
\(Rank (f_2 \circ f_1 \circ f_0) = Rank (f_1)\)
//
2: Note
The codomain of \(f_0\) is \(V_1\) with the vectors space structure of \(V_1\) and \(V_2\) is a vectors subspace of \(V'_2\) (not just a subset), which are the points.
Refer to the proposition that for any linear map between any vectors spaces and any 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have the rank of the linear map.
3: Proof
Whole Strategy: Step 1: see that \(f_2 \circ f_1 \circ f_0\) is linear; Step 2: see that \(Ran (f_1 \circ f_0) = Ran (f_1)\); Step 3: see that \(f_2 \vert_{Ran (f_1)}: Ran (f_1) \to f_2 (Ran (f_1))\) is a 'vectors spaces - linear morphisms' isomorphism, see that \(f_2 \circ f_1 \circ f_0 = f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0\), and see that \(Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_1 \circ f_0))\).
Step 1:
\(f_2 \circ f_1 \circ f_0\) is linear, which is because the codomain of \(f_0\) is the domain of \(f_1\) (more generally, a vectors subspace of the domain of \(f_1\)) and the codomain of \(f_1\) is a vectors subspace of the domain of \(f_2\): just \(V_1 \subseteq V_1\) and \(V_2 \subseteq V'_2\) sets-wise does not guarantee the linearity.
We needed to check the linearity, because otherwise, \(Rank (f_2 \circ f_1 \circ f_0)\) would not be defined.
Step 2:
Let us see that \(Ran (f_1 \circ f_0) = Ran (f_1)\).
For each \(v \in Ran (f_1 \circ f_0)\), \(v = f_1 (f_0 (v_0))\) for a \(v_0 \in V_0\), but \(v_1 := f_0 (v_0) \in V_1\) and \(v = f_1 (v_1) \in Ran (f_1)\).
For each \(v \in Ran (f_1)\), \(v = f_1 (v_1)\) for a \(v_1 \in V_1\), but as \(f_0\) is surjective, there is a \(v_0 \in V_0\) such that \(v_1 = f_0 (v_0)\), so, \(v = f_1 (v_1) = f_1 (f_0 (v_0)) \in Ran (f_1 \circ f_0)\).
That implies that \(Rank (f_1 \circ f_0) = Dim (Ran (f_1 \circ f_0)) = Dim (Ran (f_1)) = Rank (f_1)\).
Step 3:
\(Ran (f_1)\) is a vectors subspace of \(V_2\), by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
\(Ran (f_1)\) is a vectors subspace of \(V'_2\), because \(V_2\) is a vectors subspace of \(V'_2\).
\(f_2 \vert_{Ran (f_1)}: Ran (f_1) \to f_2 (Ran (f_1))\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.
\(f_2 \circ f_1 \circ f_0 = f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0\).
So, \(Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0))\).
But as \(Ran (f_1 \circ f_0) = Ran (f_1)\), \(Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0) = Ran (f_2 \vert_{Ran (f_1)})\), so, \(Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)}))\), but \(Dim (Ran (f_2 \vert_{Ran (f_1)})) = Dim (Ran (f_1))\), by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain, so, \(Rank (f_2 \circ f_1 \circ f_0) = Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0)) = Dim (Ran (f_1)) = Rank (f_1)\).
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 for linear map between vectors spaces and 'vectors spaces - linear morphisms' isomorphism, composition of linear map after isomorphism does not necessarily have rank of linear map
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 for a linear map between some vectors spaces and a 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have the rank of the 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_1\): \(: V'_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(V_0\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\) such that \(V_1 \subseteq V'_1\)
\(f_0\): \(: V_0 \to V_1\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
Statements:
Not necessarily \(Rank (f_1 \circ f_0) = Rank (f_1)\)
//
2: Note
\(V_1\) is not presupposed to be a vectors subspace of \(V'_1\) and \(V_1\) is not presupposed to equal \(V'_1\) (the composition is valid if \(V_1 \subseteq V'_1\)), which are the points.
This proposition is a warning not to jump to the conclusion without checking some things: refer to the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any vectors superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.
3: Proof
Whole Strategy: Step 1: see that \(f_1 \circ f_0\) is not guaranteed to be linear; Step 2: see a counterexample that even if \(V_1\) is a vectors subspace of \(V'_1\), \(Rank (f_1 \circ f_0) \lt Rank (f_1)\).
Step 1:
\(V_1 \subseteq V'_1\) means just that \(V_1\) is a subset of \(V'_1\) not that \(V_1\) is a vectors subspace of \(V'_1\).
For each \(v, v' \in V_0\) and each \(r, r' \in F\), \(f_1 \circ f_0 (r v + r' v') = f_1 (f_0 (r v + r' v')) = f_1 (r f_0 (v) + r' f_0 (v'))\), but the issue here is that \(r f_0 (v) + r' f_0 (v')\) is by the vectors space structure of \(V_1\) not of \(V'_1\), so, \(= r f_1 (f_0 (v)) + r' f_1 (f_0 (v'))\) is not guaranteed.
So, \(f_1 \circ f_0\) is not guaranteed to be linear.
Without being linear, \(Rank (f_1 \circ f_0)\) is not defined.
Step 2:
Let us suppose that \(V_1\) is a vectors subspace of \(V'_1\).
Let \(V'_1 = \mathbb{R}^2\) as the Euclidean vectors space, \(V_2 = \mathbb{R}^2\) as the Euclidean vectors space, \(f_1 = id_{\mathbb{R}^2}: V'_1 \to V_2\) as the identity map, \(V_0 = \mathbb{R} \oplus \{0\}\) where \(\mathbb{R}\) is the Euclidean vectors space, \(V_1 = \mathbb{R} \oplus \{0\}\) where \(\mathbb{R}\) is the Euclidean vectors space, and \(f_0 = id_{\mathbb{R} \oplus \{0\}}: V_0 \to V_1\) as the identity map.
Certainly, \(V_1 \subseteq V'_1\), where \(V_1\) is 1-dimensional, because for example, \(\{(1, 0)\}\) is a basis.
Certainly, \(f_0\) is a 'vectors spaces - linear morphisms' isomorphism.
But \(Ran (f_1) = \mathbb{R}^2\) and \(Rank (f_1) = Dim (Ran (f_1)) = 2\), while \(Ran (f_1 \circ f_0) = \mathbb{R} \otimes \{0\}\) and \(Rank (f_1 \circ f_0) = Dim (Ran (f_1 \circ f_0)) = 1\).
So, \(Rank (f_1 \circ f_0) \lt Rank (f_1)\).
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 for 'vectors spaces - linear morphisms' isomorphism, its restriction on subspace domain and corresponding range codomain is 'vectors spaces - linear morphisms' isomorphism
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 for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V'_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f'\): \(: V'_1 \to V'_2\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
\(V_1\): \(\in \{\text{ the vectors subspaces of } V'_1\}\)
\(V_2\): \(= f' (V_1)\)
\(f\): \(= f \vert_{V_1}: V_1 \to V_2\)
//
Statements:
\(f \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(V_2\) is a vectors subspace of \(V'_2\); Step 2: see that \(f\) is linear; Step 3: see that \(f\) is bijective; Step 4: conclude the proposition.
Step 1:
\(V_2 := f' (V_1)\) is a vectors subspace of \(V'_2\), by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain, which means that \(V_2\) is an \(F\) vectors space.
So, \(f\) is a map from an \(F\) vectors space into an \(F\) vectors space.
Step 2:
\(f\) is linear, because for each \(v_1, v_2 \in V_1\) and each \(r_1, r_2 \in F\), \(f (r_1 v_1 + r_2 v_2) = f' (r_1 v_1 + r_2 v_2) = r_1 f' (v_1) + r_2 f' (v_2) = r_1 f (v_1) + r_2 f (v_2)\).
Step 3:
\(f\) is bijective, because for each \(v_1, v_2 \in V_1\) such that \(v_1 \neq v_2\), \(f (v_1) = f' (v_1) \neq f' (v_2) = f (v_2)\), because \(f'\) is injective, and \(f\) is obviously surjective.
Step 4:
By the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism, \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
References
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description/proof of that for linear map between finite-dimensional vectors spaces, rank of map is rank of representative matrix w.r.t. bases
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Main Body
References
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description/proof of that determinant of square matrix over field is nonzero iff set of columns or rows is linearly independent
Topics
About:
matrices 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 determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(M\): \(\in \{\text{ the } n \times n F \text{ matrices }\}\)
//
Statements:
(
\(det M \neq 0\)
\(\iff\)
\(\{\text{ the columns of } M\} \in \{\text{ the linearly independent subsets of the } n \text{ -dimensional } F \text{ columns vectors space }\}\)
)
\(\land\)
(
\(det M \neq 0\)
\(\iff\)
\(\{\text{ the rows of } M\} \in \{\text{ the linearly independent subsets of the } n \text{ -dimensional } F \text{ rows vectors space }\}\)
)
//
2: Proof
Whole Strategy: apply Cramer's rule for any system of linear equations; Step 1: take \(M (c^1, ..., c^n)^t = 0\) and conclude for the columns; Step 2: take \(M^t (c^1, ..., c^n)^t = 0\) and conclude for the rows.
Step 1:
Let the \(j\)-th column of \(M\) be denoted as \(M_j\).
Let us take \(M (c^1, ..., c^n)^t = 0\).
It is nothing but \(c^1 M_1 + ... + c^n M_n = 0\).
So, the columns' being linearly independent is nothing but \(M (c^1, ..., c^n)^t = 0\)'s having only the 0 solution.
By Cramer's rule for any system of linear equations, that is nothing but \(det M \neq 0\): if \(det M = 0\), the rank would be smaller than \(n\), and at least 1 of \(c^j\) s could be taken arbitrarily.
So, \(det M \neq 0\) if and only if the set of the columns is linearly independent.
Step 2:
By Step 1, \(det M^t \neq 0\) if and only if the set of the columns is linearly independent.
But the set of the columns of \(M^t\) is linearly independent if and only if the set of the rows of \(M\) is linearly independent, obviously.
On the other hand, \(det M^t = det M\).
So, \(det M \neq 0\) if and only if the set of the rows of \(M\) is linearly independent.
References
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description/proof of that for finite-dimensional columns or rows vectors space and linearly independent subset, subset can be shrunk into number-of-elements-dimensional columns or rows vectors space by choosing components
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 for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(d'\): \(\in \mathbb{N}\)
\(V'_c\): \(= \{(r^1, ..., r^{d'})^t \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V'_r\): \(= \{(r^1, ..., r^{d'}) \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(S'_c\): \(= \{({r_1}^1, ..., {r_1}^{d'})^t, ..., ({r_n}^1, ..., {r_n}^{d'})^t\}\), \(\in \{\text{ the linearly independent subsets of } V'_c\}\)
\(S'_r\): \(= \{({r_1}^1, ..., {r_1}^{d'}), ..., ({r_n}^1, ..., {r_n}^{d'})\}\), \(\in \{\text{ the linearly independent subsets of } V'_r\}\)
\(V_c\): \(= \{(r^1, ..., r^n)^t \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_r\): \(= \{(r^1, ..., r^n) \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
//
Statements:
\(\exists \{j_1, ..., j_n\} \subseteq \{1, ..., d'\} (S_c := \{({r_1}^{j_1}, ..., {r_1}^{j_n})^t, ..., ({r_n}^{j_1}, ..., {r_n}^{j_n})^t\} \in \{\text{ the linearly independent subsets of } V_c\})\)
\(\land\)
\(\exists \{j_1, ..., j_n\} \subseteq \{1, ..., d'\} (S_r := \{({r_1}^{j_1}, ..., {r_1}^{j_n}), ..., ({r_n}^{j_1}, ..., {r_n}^{j_n})\} \in \{\text{ the linearly independent subsets of } V_r\})\)
//
2: Note
It is not that any \(\{j_1, ..., j_n\} \subseteq \{1, ..., d'\}\) would do: for example, when \(F = \mathbb{R}\), \(d' = 3\), and \(S'_c = \{(0, 1, 0)^t, (0, 0, 1)^t\}\), \(\{1, 2\} \subseteq \{1, 2, 3\}\) does not do because \(\{(0, 1)^t, (0, 0)^t\}\) is not linearly independent, while \(\{2, 3\} \subseteq \{1, 2, 3\}\) does.
3: Proof
Whole Strategy: Step 1: take \(c_1 ({r_1}^1, ..., {r_1}^{d'})^t + ... + c_n ({r_n}^1, ..., {r_n}^{d'})^t = 0\) and see that that equals \(\begin{pmatrix} {r_1}^1 & ... & {r_n}^1 \\ ... \\ {r_1}^{d'} & ... & {r_n}^{d'} \end{pmatrix} \begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\); Step 2: apply Cramer's rule for any system of linear equations to conclude that the rank of the matrix is \(n\); Step 3: choose a linearly independent \(S_c\); Step 4: conclude for \(S_r\).
Step 1:
Let us take \(c_1 ({r_1}^1, ..., {r_1}^{d'})^t + ... + c_n ({r_n}^1, ..., {r_n}^{d'})^t = 0\).
That equals \(\begin{pmatrix} {r_1}^1 & ... & {r_n}^1 \\ ... \\ {r_1}^{d'} & ... & {r_n}^{d'} \end{pmatrix} \begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\), obviously.
\(S_c\)' s being linearly independent equals that equation's having only the \(\begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\) solution.
Step 2:
By Cramer's rule for any system of linear equations, the rank of the matrix is \(n\): if the rank was smaller than \(n\), at least 1 of \(c_j\) s can be taken arbitrarily, a contradiction.
Step 3:
So, the matrix can be rearranged such that the top-left \(n \times n\) submatrix is determinant nonzero.
Then, the \(n\) columns of the submatrix is linearly independent, by the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
The \(n\) columns is an \(S_c\).
Step 4:
As for \(S_r\), we can just take the transpose of \(S'_r\), \(S'_c \subseteq V'_c\), take an \(S_c\), and take the transpose of \(S_c\), \(S_r \subseteq V_r\).
References
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