Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $W_1$ be a subspace of a finite dimensional vector space $V$. Prove that there exists a subspace $W_2\subset V$ such that $V=W_1\oplus W_2$.

EDIT$^1$:

This may be of use here:

enter image description here enter image description here


What does $$ W_1 \cap W_2 = \{0\}$$ even mean? The zero vector is the only thing shared in common by the two subset?

share|improve this question
4  
I don't see how this "could mean various things". Can you elaborate on where you think the ambiguity is? –  Alex Becker May 22 '13 at 5:47
    
$W_1$ could be the set $\{0\}$, and so could $W_2$, right? –  Trancot May 22 '13 at 5:49
    
In that case, if $V$ is not trivial, then $W_1 + W_2 \neq V$. –  Adam Saltz May 22 '13 at 5:51
    
I mean it could be, so it's hard to generalize, right? –  Trancot May 22 '13 at 5:52
4  
I'm not sure what any of your questions are asking, to be honest. I think you should review the definitions of the terms and symbols in the question. BTW, you are correct that $W_1\cap W_2=\{0\}$ means that the only thing shared by the two subspaces is the zero vector. –  Alex Becker May 22 '13 at 6:02
show 9 more comments

3 Answers 3

up vote 4 down vote accepted

Choose a basis for $W_1$. This basis can be extended to a basis for $V$ (well-known result). Now let $W_2$ be the span of the newly added vectors. Then $W_1 \cap W_2 = \{0\}$ by linear independence and $W_1 + W_2 = V$.

share|improve this answer
    
Yes, Jesus, that is precisely what is stated in Cohn's work under the name complement, right? –  Trancot May 22 '13 at 6:59
    
Yeah, seems so. –  Jesus May 22 '13 at 7:01
    
Ah, you beat me to it. Deleted my answer. (+1) –  Bryan Urízar May 22 '13 at 7:03
add comment

Consider the set $\mathcal W$ of subspaces $W\subseteq V$ with $W_1\cap W=\{0\}$. Let $W_2$ be a maximal element of $\mathcal W$ (with respect to inclusion). Why does such maximal element exist and why does that solve the problem?

share|improve this answer
add comment

Here are some explicit examples for you to play around with:

  • Let $V$ be 2-dimensional, say $\langle x, y\rangle$, and let $W_1 = \langle x\rangle$. Can you see that $W_2 = \langle y\rangle$ works? Can you see that $\langle x+y\rangle$ also works, and that $\langle 3x - 5y \rangle$ also works, but that $\langle x \rangle$ doesn't, that $\{0\}$ doesn't, and that $\langle x+y, x-y\rangle$ doesn't? (Draw pictures.)
  • Let $V = \langle x_1, \dots, x_n\rangle$ (n large), $W_1 = \langle x_1, x_2, x_3\rangle$, and show that $W_2 = \langle x_4, \dots, x_n\rangle$ works.
  • Let $V = \langle x, y, z\rangle$, and let $W_1 = \langle x\rangle$. Can you pick the obvious space that works as $W_2$? Can you pick a non-obvious space that works? Can you show why $\langle y\rangle$ and $\langle 3x+5y, x-y\rangle$ don't work?
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.