description/proof of that projection from vectors space into subspace w.r.t. complementary subspace is linear map and image of any subspace is 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
Target Context
- The reader will have a description and a proof of 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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(\widetilde{V}\): \(\in \{\text{ the complementary subspaces of } V\}\)
\(f\): \(: V' \to V\), \(= \text{ the projection map into } V \text{ with respect to } \widetilde{V}\)
\(V''\): \(\in \{\text{ the vectors subspaces of } V'\}\)
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Statements:
\(f \in \{\text{ the linear maps }\}\)
\(\land\)
\(f (V'') \in \{\text{ the subspaces of } V'\}\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V'\), any vectors subspace of \(V'\), \(V\), any complementary subspace of \(V\), \(\widetilde{V}\), the projection map into \(V\) with respect to \(\widetilde{V}\), \(f: V' \to V\), and any vectors subspace, \(V'' \subseteq V'\), \(f\) is a linear map, and \(f (V'')\) is a vectors subspace of \(V'\).
3: Proof
Whole Strategy: Step 1: choose any \(v'_1, v'_2 \in V'\) and any \(r_1, r_2 \in F\), and see that \(f (r_1 v'_1 + r_2 v'_2) = r_1 f (v'_1) + r_2 f (v'_2)\); Step 2: choose any \(v_1, v_2 \in f (V'')\), and see that \(r_1 v_1 + r_2 v_2 \in f (V'')\).
Step 1:
For any \(v'_1, v'_2 \in V'\) and any \(r_1, r_2 \in F\), \(f (r_1 v'_1 + r_2 v'_2) = r_1 f (v'_1) + r_2 f (v'_2)\)?
\(v'_j = v_j + \widetilde{v}_j\) where \(v_j \in V\) and \(\widetilde{v}_j \in \widetilde{V}\). \(r_1 v'_1 + r_2 v'_2 = r_1 (v_1 + \widetilde{v}_1) + r_2 (v_2 + \widetilde{v}_2) = r_1 v_1 + r_2 v_2 + r_1 \widetilde{v}_1 + r_2 \widetilde{v}_2\).
\(f (r_1 v'_1 + r_2 v'_2) = r_1 v_1 + r_2 v_2 = r_1 f (v'_1) + r_2 f (v'_2)\).
So, yes.
Step 2:
For any \(v_1, v_2 \in f (V'')\), \(r_1 v_1 + r_2 v_2 \in f (V'')\)?
There is a \(v''_j \in V''\) such that \(v_j = f (v''_j)\). \(r_1 v''_1 + r_2 v''_2 \in V''\). \(f (r_1 v''_1 + r_2 v''_2) = r_1 f (v''_1) + r_2 f (v''_2) = r_1 v_1 + r_2 v_2 \in f (V'')\).
So, yes.
