description/proof of that for vectors space and 2 same-finite-dimensional vectors subspaces, there is common complementary 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 complementary subspace of vectors subspace.
- The reader admits the proposition that for any vectors space, the intersection of any possibly uncountable number of finite-dimensional subspaces is a subspace with a dimension equal to or smaller than the minimum dimension of the subspaces.
- The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.
- 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 and any 2 same-finite-dimensional subspaces, there is a common 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_1\): \(\in \{\text{ the } k \text{ -dimensional subspaces of } V\}\)
\(V_2\): \(\in \{\text{ the } k \text{ -dimensional subspaces of } V\}\)
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Statements:
\(\exists V_3 \in \{\text{ the subspaces of } V\} (V_3 \in \{\text{ the complementary subspaces of } V_1\} \cap \{\text{ the complementary subspaces of } V_2\})\)
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2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V\), and any \(k\)-dimensional subspaces, \(V_1, V_2\), there is a common complementary subspaces of \(V_1\) and \(V_2\), \(V_3 \subseteq V\).
3: Proof
Whole Strategy: we are going to explicitly construct a \(V_3\); Step 1: take any basis of \(V_1 \cap V_2\) as \(\{e_1, ..., e_l\}\); Step 2: add to \(\{e_1, ..., e_l\}\) some elements of \(V_j\) to form a basis of \(V_j\), \(\{e_1, ..., e_l, e_{j, l + 1}, ..., e_{j, k}\}\); Step 3: see that \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\) is linearly independent; Step 4: add to the bases of \(V_1\) and \(V_2\) \(\{e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\) and see that the results are linearly independent and span the same subspace spanned by \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\); Step 5: add to \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\) some elements of \(V\), \(\{e'_1, e'_2, ...\}\), to form a basis of \(V\); Step 6: define \(V_3\) as the space spanned by \(\{e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\); Step 7: see that \(V_3\) is indeed a common complementary subspace of \(V_1\) and \(V_2\).
Let us construct a common complementary subspace.
Step 1:
\(V_1 \cap V_2\) is a subspace of a dimension, \(l \le k\), by the proposition that for any vectors space, the intersection of any possibly uncountable number of finite-dimensional subspaces is a subspace with a dimension equal to or smaller than the minimum dimension of the subspaces. Let us take any basis of \(V_1 \cap V_2\), \(\{e_1, ..., e_l\}\). \(l\) may be \(0\), and in that case, it becomes the empty set.
Step 2:
\(\{e_1, ..., e_l\}\) is linearly independent on \(V_1\), because otherwise, \(\sum_{j \in \{1, ..., l\}} c^j e_j = 0\) would have a not-all-zero \((c^1, ..., c^l)\), which would mean that \(\{e_1, ..., e_l\}\) was linearly dependent on \(V_1 \cap V_2\). Likewise, \(\{e_1, ..., e_l\}\) is linearly independent on \(V_2\).
Let us add some \(k - l\) elements of \(V_1\), \(\{e_{1, l + 1}, ..., e_{1, k}\}\), to \(\{e_1, ..., e_l\}\) to form a basis for \(V_1\) and add some \(k - l\) elements of \(V_2\), \(\{e_{2, l + 1}, ..., e_{2, k}\}\), to \(\{e_1, ..., e_l\}\) to form a basis for \(V_2\), which is possible by the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.
Step 3:
\(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\) is linearly independent on \(V\), because if a \(e_{2, l + j}\) was a linear combination of the other elements as \(e_{2, l + j} = \sum_{m \in \{1, ..., l\}} c^m e_m + \sum_{n \in \{1, ..., k - l\}} c'^n e_{1, l + n} + \sum_{o \in \{1, ..., j - 1, j + 1, ..., k - l\}} c''^o e_{2, l + o}\), \(e_{2, l + j} - \sum_{o \in \{1, ..., j - 1, j + 1, ..., k - l\}} c''^o e_{2, l + o} = \sum_{m \in \{1, ..., l\}} c^m e_m + \sum_{n \in \{1, ..., k - l\}} c'^n e_{1, l + n}\), but the left hand side would not be \(0\) because \((e_{2, l + 1}, ..., e_{2, k})\) would be linearly independent, and the right hand side would be a vector on \(V_1\), and so, \(e_{2, l + j} - \sum_{o \in \{1, ..., j - 1, j + 1, ..., k - l\}} c''^o e_{2, l + o}\) would be in \(V_1 \cap V_2\), which would imply that it would be a linear combination of \(e_1, e_2, ..., e_l\), a contradiction against \(\{e_1, ..., e_l, e_{2, l + 1}, ..., e_{2, k}\}\)'s being linearly independent.
Step 4:
Let us add \(e_{1, l + 1} + e_{2, l + 1}\) to the both bases. Then, the both results are linearly independent, because for \(V_1\), \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}\}\) is linearly independent from the fact in the previous paragraph (for \(\sum_{m \in \{1, ..., l\}} c^m e_m + \sum_{n \in \{1, ..., k - l\}} c'^n e_{1, l + n} + c'''_1 (e_{1, l + 1} + e_{2, l + 1}) = \sum_{m \in \{1, ..., l\}} c^m e_m + (c'^1 + c'''_1) e_{1, l + 1} + \sum_{n \in \{2, ..., k - l\}} c'^n e_{1, l + n} + c'''_1 e_{2, l + 1}) = 0\), \(c'''_1 = 0\) and \(c'^1 + c'''_1 = 0\), so, \(c'^1 = 0\), and all the other coefficients are \(0\)), and likewise for \(V_2\).
Then, let us add \(e_{1, l + 2} + e_{2, l + 2}\) to the both results of the previous paragraph. Then, the both results are linearly independent, because for \(V_1\), \(\{e_1, e_2, ..., e_l, e_{1, l + 1}, e_{1, l + 2}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, e_{1, l + 2} + e_{2, l + 2}\}\) is linearly independent (for \(\sum_{m \in \{1, ..., l\}} c^m e_m + \sum_{n \in \{1, ..., k - l\}} c'^n e_{1, l + n} + c'''_1 (e_{1, l + 1} + e_{2, l + 1}) + c'''_2 (e_{1, l + 2} + e_{2, l + 2}) = \sum_{m \in \{1, ..., l\}} c^m e_m + (c'^1 + c'''_1) e_{1, l + 1} + (c'^2 + c'''_2) e_{1, l + 2} + \sum_{n \in \{3, ..., k - l\}} c'^n e_{1, l + n} + c'''_1 e_{2, l + 1} + c'''_2 e_{2, l + 2}) = 0\), \(c'''_1, c'''_2 = 0\) and \(c'^1 + c'''_1, c'^2 + c'''_2 = 0\), so, \(c'^1, c'^2 = 0\), and all the other coefficients are \(0\)), and likewise for \(V_2\).
And so on until we have added \(e_{1, k} + e_{2, k}\). Then, the both results span the same subspace, the space spanned by \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\), denoted as \(V_4\). That is because while each element of \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\) is a linear combination of \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\), each element of \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\) is a linear combination of \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\): \(e_{2, j} = e_{1, j} + e_{2, j} - e_{1, j}\), and likewise for \(\{e_1, ..., e_l, e_{2, l + 1}, ..., e_{2, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\). \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}\}\), \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\), and \(\{e_1, ..., e_l, e_{2, l + 1}, ..., e_{2, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}\}\) are some bases of \(V_4\).
Step 5:
Let us add some vectors, \(\{e'_1, e'_2, ...\}\) (although indexed like that here for convenience of expression, they may be uncountably many vectors), to the 1st basis of \(V_4\) to form a basis of \(V\), \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{2, l + 1}, ..., e_{2, k}, e'_1, e'_2, ...\}\), 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. In fact, obviously, also \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\) and \(\{e_1, ..., e_l, e_{2, l + 1}, ..., e_{2, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\) are some bases of \(V\).
Step 6:
Then, we define \(V_3\) as the space spanned by \(\{e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\).
Step 7:
Let us see that \(V_3\) is indeed a complementary subspace of \(V_1\).
Let us see that \(V = V_1 + V_3\). \(V\) is spanned by \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\). \(V_1\) is spanned by \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}\}\). \(V_3\) is spanned by \(\{e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\).
Let us see that \(V_1 \cap V_3 = \{0\}\). For any nonzero element of \(V_3\), \(c^1 (e_{1, l + 1} + e_{2, l + 1}) + ... + c^k (e_{1, k} + e_{2, k}) + c'^1 e'_1 + c'^2 e'_2 + ...\), it cannot be any linear combination of the basis of \(V_1\), \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}\}\), because the 2nd basis of \(V\), \(\{e_1, ..., e_l, e_{1, l + 1}, ..., e_{1, k}, e_{1, l + 1} + e_{2, l + 1}, ..., e_{1, k} + e_{2, k}, e'_1, e'_2, ...\}\) is linearly independent, which means that it is not any element of \(V_1\).
Likewise, \(V_3\) is indeed a complementary subspace of \(V_2\).
It is clear that it does not matter if \(l = 0\) or \(l = k\) or \(k\) equals \(0\) or the dimension of \(V\). In fact, if \(l = 0\), we can do just without \(\{e_1, ..., e_l\}\); if \(l = k\), the 2 subspaces are the same, and there is a common complementary subspace anyway; if \(k = 0\), the 2 subspaces are the same \(\{0\}\) and the whole \(V\) is the common complementary subspace; if \(k\) equals the dimension of \(V\), the 2 subspaces are the same \(V\) and \(\{0\}\) is the common complementary subspace.
