722: For Vectors Space and 2 Same-Finite-Dimensional Vectors Subspaces, There Is Common Complementary Subspace

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description/proof of that for vectors space and 2 same-finite-dimensional vectors subspaces, there is common complementary subspace

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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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.

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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.

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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.

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Main Body

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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:
$\mathrm{\exists }{V}_3\in \{\text{ the subspaces of }V\}(V_3\in \{\text{ the complementary subspaces of }{V}_1\}\cap \{\text{ the complementary subspaces of }{V}_2\})$
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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 $V_2$, $V_3\subseteq V$.

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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 $V$, $\{e_1^{\prime },e_2^{\prime },...\}$, to form a basis of $V$; Step 6: define $V_3$ as the space spanned by $\{e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$; 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 $e_{2,l+j}=\sum _{m\in \{1,...,l\}}{c}^m{e}_m+\sum _{n\in \{1,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}+\sum _{o\in \{1,...,j-1,j+1,...,k-l\}}{c}^{″o}{e}_{2,l+o}$, $e_{2,l+j}-\sum _{o\in \{1,...,j-1,j+1,...,k-l\}}{c}^{″o}{e}_{2,l+o}=\sum _{m\in \{1,...,l\}}{c}^m{e}_m+\sum _{n\in \{1,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}$, 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 $\sum _{m\in \{1,...,l\}}{c}^m{e}_m+\sum _{n\in \{1,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}+c_1^{‴}(e_{1,l+1}+e_{2,l+1})=\sum _{m\in \{1,...,l\}}{c}^m{e}_m+(c^{\prime 1}+c_1^{‴})e_{1,l+1}+\sum _{n\in \{2,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}+c_1^{‴}{e}_{2,l+1})=0$, $c_1^{‴}=0$ and $c^{\prime 1}+c_1^{‴}=0$, so, $c^{\prime 1}=0$, 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 $\sum _{m\in \{1,...,l\}}{c}^m{e}_m+\sum _{n\in \{1,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}+c_1^{‴}(e_{1,l+1}+e_{2,l+1})+c_2^{‴}(e_{1,l+2}+e_{2,l+2})=\sum _{m\in \{1,...,l\}}{c}^m{e}_m+(c^{\prime 1}+c_1^{‴})e_{1,l+1}+(c^{\prime 2}+c_2^{‴})e_{1,l+2}+\sum _{n\in \{3,...,k-l\}}{c}^{\prime n}{e}_{1,l+n}+c_1^{‴}{e}_{2,l+1}+c_2^{‴}{e}_{2,l+2})=0$, $c_1^{‴},c_2^{‴}=0$ and $c^{\prime 1}+c_1^{‴},c^{\prime 2}+c_2^{‴}=0$, so, $c^{\prime 1},c^{\prime 2}=0$, 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, $\{e_1^{\prime },e_2^{\prime },...\}$ (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 $V$, $\{e_1,...,e_l,e_{1,l+1},...,e_{1,k},e_{2,l+1},...,e_{2,k},e_1^{\prime },e_2^{\prime },...\}$, 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 $\{e_1,...,e_l,e_{1,l+1},...,e_{1,k},e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$ and $\{e_1,...,e_l,e_{2,l+1},...,e_{2,k},e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$ are some bases of $V$.

Step 6:

Then, we define $V_3$ as the space spanned by $\{e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$.

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 $\{e_1,...,e_l,e_{1,l+1},...,e_{1,k},e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$. $V_1$ is spanned by $\{e_1,...,e_l,e_{1,l+1},...,e_{1,k}\}$. $V_3$ is spanned by $\{e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$.

Let us see that $V_1\cap V_3=\{0\}$. For any nonzero element of $V_3$, $c^1(e_{1,l+1}+e_{2,l+1})+...+c^k(e_{1,k}+e_{2,k})+c^{\prime 1}{e}_1^{\prime }+c^{\prime 2}{e}_2^{\prime }+...$, 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 $V$, $\{e_1,...,e_l,e_{1,l+1},...,e_{1,k},e_{1,l+1}+e_{2,l+1},...,e_{1,k}+e_{2,k},e_1^{\prime },e_2^{\prime },...\}$ 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.

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References

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