Common supplementary subspaces

[geoffrey159]

Homework Statement

Let $E$ be a finite dimensional vector space, $A$ and $B$ two subspaces with the same dimension.
Show there is a subspace $S$ of $E$ such that $E = A \bigoplus S = B \bigoplus S$

Homework Equations

[/B]
$\text{dim}(E) = n$
$\text{dim}(A) = \text{dim}(B) = m \le n$

${\cal B} = (e_1,...,e_n)$ is a basis of $E$
$A = \text{span}(e_{i_1},...,e_{i_m})$
$B = \text{span}(e_{j_1},...,e_{j_m})$

$S_A = \text{span}((e_i)_{i \neq i_1,...,i_m } )$
$S_B = \text{span}((e_i)_{i \neq j_1,...,j_m } )$

$E = A \bigoplus S_A = B \bigoplus S_B$

The Attempt at a Solution

I find this exercise hard and I can't finish it. Could you help please ?The case $A = B$ is easy, $S = S_A = S_B$

I assume now that $A\neq B$. I want to show the result by induction based on the decreasing dimensions of $A$ and $B$.
1 - If $m = n$ then $A = B = E$ and $S = \{0\}$ works
2 - If A and B are hyperplanes $m = n-1$, then $S_A = \text{span}(e_k)$, and $S_B = \text{span}(e_\ell)$, with $k\neq \ell$.
Put $S = \text{span}(e_k + e_\ell)$.

- Then for any $x \in A\cap S$, there are scalars $(\lambda_i)_{i\neq k}$ and $\mu$ such that $x = \sum_{i\neq k } \lambda_i e_{i} = \mu (e_k + e_\ell ) \Rightarrow 0 = \mu (e_k + e_\ell ) - \sum_{i\neq k } \lambda_i e_{i} = \mu e_k + ( \mu - \lambda_l) e_l - \sum_{i\neq k,l } \lambda_i e_{i}$. Since ${\cal B}$ is a basis of $E$, then $\mu = 0$ and $x = 0$. So $A \cap S =\{0\}$. We can do the same for $B$ be so that $B\cap S=\{0\}$
- For any $x \in E$, there are scalars $(\lambda_i)_{i = 1...n}$ such that $x = \sum_{i= 1}^n \lambda_i e_i$. Reordering the terms,
$x = \lambda_k (e_k + e_\ell) + ((\lambda_\ell - \lambda_k) e_\ell + \sum_{i\neq k,\ell}^n \lambda_i e_i) \Rightarrow x \in S + A \Rightarrow S + A = E$
$x = \lambda_l (e_k + e_\ell) +( (\lambda_k - \lambda_\ell) e_k + \sum_{i\neq k,\ell}^n \lambda_i e_i) \Rightarrow x \in S + B \Rightarrow S + B = E$​

The two points above show that $E = A \bigoplus S = B \bigoplus S$
3 - Assume that it works for $m = n,n-1,...,r+1$. I want to show it works for $m = r$

- If $S_A \cap S_B \neq \emptyset$, then there is a vector $e_k \in S_A \cap S_B$ such that $S_A = \text{span}(e_k)\bigoplus S_A'$, and $S_B = \text{span}(e_k)\bigoplus S_B'$, where $S'_A = \text{span}((e_i)_{i \neq k,i_1,...,i_m } )$
and $S'_B = \text{span}((e_i)_{i \neq k,j_1,...,j_m } )$
So $E = (A \bigoplus \text{span}(e_k)) \bigoplus S_A' = (B \bigoplus \text{span}(e_k)) \bigoplus S_B'$. By induction hypothesis, there is a subspace $S'$ such that :
$E = (A \bigoplus \text{span}(e_k)) \bigoplus S' = (B \bigoplus \text{span}(e_k)) \bigoplus S'$. So $S = \text{span}(e_k) \bigoplus S'$ works.
- If $S_A \cap S_B = \emptyset$, I don't know how to finish that part !
​

 

[geoffrey159]

Point 3.1 does not convince me anymore, I need to rework that ! Sorry.
And $S_A \cap S_B \neq \emptyset$, always, because the intersection of subspaces is a subspace, so it contains 0.

 

[geoffrey159]

I think I might have more. Point 2 was more important than I thought.

Any finite dimensional vector space can be broken into the direct sum of an hyperplane and a vectorial line : $E = H \bigoplus D$

Point 2 on OP shows that if $E = H_1 \bigoplus D_1 = H_2 \bigoplus D_2$, then there exist a vectorial line $D$ such that $E = H_1 \bigoplus D = H_2 \bigoplus D$.

So now assume that the dimension of $S_A$ and $S_B$ is 2. They can both be broken into a direct sum of an hyperplane and a vectorial line, so :
$E = A \bigoplus S_A = A \bigoplus ( H_1 \bigoplus D_1) = (A \bigoplus H_1) \bigoplus D_1$,
$E = B \bigoplus S_B = B \bigoplus ( H_2 \bigoplus D_2) =(B \bigoplus H_2) \bigoplus D_2$.
We have that $A \bigoplus H_1$ and $B \bigoplus H_2$ are hyperplanes of $E$, so there is a line $D$ such that
$E = (A \bigoplus H_1) \bigoplus D = (B \bigoplus H_2)\bigoplus D$. Commute $H_1$ and $D$, as well as $H_2$ and $D$, and do the same. At the end, $E = A \bigoplus (D' \bigoplus D) = B \bigoplus (D' \bigoplus D)$

Repeat this operation by induction. Is this idea good ?

 

[HallsofIvy]

If A is a subspace of E then, given any basis for A, b= {v1, v2, ..., vm}, there exist a basis for E that contains b.

 

[geoffrey159]

Thank you for your reply. I really need some feedback on this exercise, which is horrible !
I understand your comment, in the general case, I should have completed a basis of A and B with vectors of the basis ${\cal B}$ of E, but to tell you the truth, I was completely helpless at first, so I took a less general case in order to see clearer. Please allow me to restate it taking into account your comment.

Let ${\cal B} = \{ e_1, ..., e_n\}$ be a basis of $E$

0 - If $A = B$, since every subspace of a finite dimensional vector space has at least one supplementary space, $A$ and $B$ share this supplementary space.

1 - If $A$ and $B$ have dimension $n$, then $A = B = E$, and with remark 0, they have a common supplementary space which is $S=\{0\}$

2 - In general, a finite dimensional vector space $W$ can be expressed as the direct sum of an hyperplane $H$ and a vectorial line $D$. If you consider a basis $(f_1,...,f_p)$ of that vector space, put $H = \text{span}(f_1,...,f_{p-1})$, and $D = \text{span}(f_p)$. It is easy to see that $W = H + D$ and $H \cap D = \{0\}$, so that $W = H \bigoplus D$

3 - If $A$ and $B$ have dimension $n-1$, they are hyperplanes of E. One can complete a basis of $A$ (respectively $B$) with one vector $e_A \in {\cal B}$ (respectively $e_B$) in order to form a new basis of $E$. Setting $D_A = \text{span}(e_A)$, $D_B = \text{span}(e_B)$, and following remark 2, then $E = A \bigoplus D_A = B \bigoplus D_B$.

4 - If $D_A = D_B$ then $D_A$ is a common supplementary space of $A$ and $B$

5 - Assume $D_A \neq D_B$, which amounts to saying that $e_A \in B$ and $e_B \in A$. Set $D = \text{span}(e_A + e_B)$. I want to show that $E = A \bigoplus D$ (respectively $E = B \bigoplus D$).

• We already know that $A + D \subset E$. For any $x\in E = A \bigoplus D_A$, it can be expressed uniquely as $x = a + \lambda e_A$, $a\in A$ and $\lambda$ scalar. So $x = (a - \lambda e_B) + \lambda (e_A + e_B) \Rightarrow x \in A + D \Rightarrow E \subset A + D$. That shows $E = A + D$.
• Now we need to show that $A\cap D = \{0\}$. If $x \in A \cap D$, then it has the form $x = \sum_{i = 1}^{n-1} \lambda_i a_i = \mu (e_A + e_B)$ when $(a_1,...,a_{n-1})$ is a basis of A. Furthermore $e_B \in A$ so $e_B = \sum_{i = 1}^{n-1} \alpha_i a_i$. It follows that $\mu e_A + \sum_{i=1}^{n-1} (\mu \alpha_i - \lambda_i) a_i = 0$. Since $(a_1,...,a_{n-1}, e_A)$ is a basis of $E$, then $\mu = 0$ and $x = 0$. That shows $A\cap D = \{0\}$
• Doing the same thing for $B$, I get what I want.

6 - Assume that if the dimension of $A$ and $B$ is greater or equal to $r + 1$, then they have a common supplementary.
Assume that now $A$ and $B$ have dimension r. Then,
$E = A \bigoplus S_A = A \bigoplus (D_A \bigoplus H_A) = (A \bigoplus D_A) \bigoplus H_A$
$E = B \bigoplus S_B = B \bigoplus (D_B \bigoplus H_B) = (B \bigoplus D_B) \bigoplus H_B$
Both $A \bigoplus D_A$ and $B \bigoplus D_B$ have dimension $r+1$, so they have a common supplementary $H$ in $E$.
Commuting the terms, $E = ( A \bigoplus H ) \bigoplus D_A = (B\bigoplus H) \bigoplus D_B$. Now $A \bigoplus H$ and $B \bigoplus H$ are hyperplanes of $E$ so they have a common supplementary $D$.
It follows that $E = A \bigoplus (H \bigoplus D) = B \bigoplus (H \bigoplus D)$. So by induction, it can be shown.

 

[geoffrey159]

I apologize for putting my thread on top of the list like that, I know it's not very polite , but I would like to know if post #5 answers the question.