description/proof of that for vectors space, subspace, and complementary subspace, finite-dimensional subspace that intersects complementary subspace trivially is projected into subspace as same-dimensional subspace
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of projection of vector into vectors subspace with respect to complementary subspace.
- The reader admits the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace.
- The reader admits the proposition that for any linear map from any finite-dimensional vectors space, there is a domain subspace that is 'vectors spaces - linear morphisms' isomorphic to the map range by the restriction of the map on the subspace domain.
- The reader admits the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension..
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space, any subspace, and any complementary subspace, any finite-dimensional subspace that intersects the complementary subspace trivially is projected into the subspace as a same-dimensional subspace.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V\): \(\in \{\text{ the subspaces of } V'\}\)
\(\widetilde{V}\): \(\in \{\text{ the complementary subspaces of } V\}\)
\(f\): \(V' \to V\), \(= \text{ the projection into } V \text{ with respect to } \widetilde{V}\)
\(V''\): \(\in \{\text{ the } d \text{ -dimensional subspaces of } V'\}\)
//
Statements:
\(V'' \cap \widetilde{V} = \{0\}\)
\(\implies\)
\(f (V'') \in \{\text{ the } d \text{ -dimensional subspaces of } V'\}\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V'\), any subspace, \(V \subseteq V'\), any complementary subspace, \(\widetilde{V} \subseteq V'\), the projection into \(V\) with respect to \(\widetilde{V}\), \(f: V' \to V\), and any d-dimensional subspace, \(V'' \subseteq V'\), such that \(V'' \cap \widetilde{V} = \{0\}\), \(f (V'')\) is a d-dimensional subspace of \(V'\).
3: Proof
Whole Strategy: Step 1: see that \(f (V'')\) is a subspace of \(V'\); Step 2: choose a subspace, \(V''' \subseteq V''\), that is 'vectors spaces - linear morphisms' isomorphic to \(f (V'')\); Step 3: suppose that the dimension of \(V'''\) was smaller than \(d\) and find a contradiction; Step 4: see that the dimension of \(f (V'')\) is \(d\).
Step 1:
\(f (V'')\) is a subspace of \(V'\), by the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace.
Step 2:
There is a subspace, \(V''' \subseteq V''\), that is 'vectors spaces - linear morphisms' isomorphic to \(f (V'')\) by the restriction of \(f\) on the \(V'''\) domain, \(f': V''' \to f (V'')\), by the proposition that for any linear map from any finite-dimensional vectors space, there is a domain subspace that is 'vectors spaces - linear morphisms' isomorphic to the map range by the restriction of the map on the subspace domain.
Step 3:
Let us suppose the dimension of \(V'''\) to be \(l\). Also the dimension of \(f (V'')\) is \(l\), by the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension..
Let us suppose that \(l \lt d\).
There would be a basis for \(V'''\), \(\{e_1, ..., e_l\}\). There would be an element, \(v \in V''\), that was not any linear combination of the basis. \(\{f' (e_1), ..., f' (e_l)\}\) would be a basis for \(f (V'')\), because \(f'\) was a 'vectors spaces - linear morphisms' isomorphism. \(f (v) = c^1 f' (e_1) + ... + c^l f' (e_l)\). Let us define \(v' = v - (c^1 e_1 + ... + c^l e_l)\), nonzero, because \(v\) was not any linear combination of the basis. \(f (v') = f (v - (c^1 e_1 + ... + c^l e_l)) = f (v) - (c^1 f (e_1) + ... + c^l f (e_l))\), because \(f\) was linear, by the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace, \(= f (v) - (c^1 f' (e_1) + ... + c^l f' (e_l)) = 0\). Then, \(v' = 0 + v'' = v''\) for a nonzero \(v'' \in \widetilde{V}\), which would imply that \(V''\) intersected \(\widetilde{V}\) non-trivially, a contradiction.
So, \(l = d\).
Step 4:
As the dimension of \(f (V'')\) is \(l\), it is \(l = d\).
