description/proof of that vectors subspace has complementary subspace
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of complementary subspace of vectors subspace.
- The reader knows a definition of vectors subspace generated by subset of vectors space.
- The reader admits the proposition that any vectors space has a basis.
- The reader admits the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space, any vectors subspace has a complementary 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V\): \(\in \{\text{ the vectors subspaces of } V'\}\)
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Statements:
\(\exists \widetilde{V} \in \{\text{ the complementary vectors subspaces of } V\}\)
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2: Note
Each of \(V'\) and \(V\) does not need to be finite-dimensional.
This proposition is not saying that \(\widetilde{V}\) is unique but saying that there is at least 1 \(\widetilde{V}\): refer to Note 2 for the definition of complementary subspace of vectors subspace.
3: Proof
Whole Strategy: Step 1: take any basis for \(V\), \(B\); Step 2: expand \(B\) to be a basis for \(V'\), \(B'\); Step 3: take \(\widetilde{B} := B' \setminus B\) and take \(\widetilde{V} := (\widetilde{B})\), and see that \(\widetilde{V}\) is a complementary subspace of \(V\).
Step 1:
Let us take any basis for \(V\), \(B = \{b_j\}\), which is possible, by the proposition that any vectors space has a basis.
Step 2:
Let us expand \(B\) to be a basis for \(V'\), \(B' = \{b'_j\}\), which is possible, by the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis: \(B\) is linearly independent on \(V'\).
Step 3:
Let us take \(\widetilde{B} := B' \setminus B = \{\widetilde{b}_j\}\).
Let us take \(\widetilde{V} := (\widetilde{B})\), which is the vectors subspace generate by \(\widetilde{B}\), which is nothing but the span of \(\widetilde{B}\): refer to Note for the definition of vectors subspace generated by subset of vectors space.
Let us see that \(\widetilde{V}\) is a complementary subspace of \(V\).
Let \(v' \in V \cap \widetilde{V}\) be any.
\(v' = \sum_{j \in J} v'^j b_j = \sum_{l \in L} \widetilde{v'}^l \widetilde{b}_l\) for some finite \(J\) and \(L\).
So, \(\sum_{j \in J} v'^j b_j - \sum_{l \in L} \widetilde{v'}^l \widetilde{b}_l = 0\), but as \(\{b_j \vert j \in J\} \cup \{\widetilde{b}_l \vert l \in L\}\) is a non-duplicate finite subset of \(B'\), it is linearly independent, and \(v'^j = 0\) and \(\widetilde{v'}^l = 0\), so, \(v' = 0\).
So, \(V \cap \widetilde{V} = \{0\}\).
Let \(v' \in V'\) be any.
\(v' = \sum_{m \in M} v'^m b'_m\) for a finite \(M\), but each \(b'_m\) is exclusively in \(B\) or \(\widetilde{B}\), so, \(\sum_{m \in M} v'^m b'_m = \sum_{j \in J} v'^j b_j + \sum_{l \in L} v'^l \widetilde{b}_l\) where \(J\) and \(L\) are finite, while \(\sum_{j \in J} v'^j b'_j \in V\) and \(\sum_{l \in L} v'^l \widetilde{b}_l \in \widetilde{V}\).
So, \(V' \subseteq V + \widetilde{V}\).
As obviously, \(\subseteq V + \widetilde{V} \subseteq V'\), \(V' = V + \widetilde{V}\).
So, \(\widetilde{V}\) is a complementary subspace of \(V\).
