3
$\begingroup$

Let W and W′ be two finite dimensional subspaces of an inner product space such that dim(W) < dim(W′). Prove that there is a nonzero vector x in W′ orthogonal to W.

This should be an easy problem but I'm stuck. Here's what I got: find a basis S for the intersection of W and W′ and then apply Gram-Schmidt. Then extend it into S′ a basis for W′ and Gram-Schmidt again. There should at least be one vector in S′ but not S. Call it x. It should be orthogonal to W. But I don't know how to show that.

OK. After some example in lower dimensional space, I found out the reason I'm stuck is because my claim is not true. We might have to do this using dimension.

Please help!

$\endgroup$
3
$\begingroup$

We can reduce the problem to a finite dimensional one by having the ambient space be $Z=W+W'$.

$\dim (W^\perp\cap Z)=\dim Z-\dim W$ and thus \begin{align}\dim (W^\perp\cap W')=&\dim W'+\dim (W^\perp\cap Z)-\dim (W'+(W^\perp\cap Z))\ge \\\ge&\dim W'+\dim (W^\perp\cap Z)-\dim Z=\dim W'-\dim W>0\end{align}

Which proves the claim.

$\endgroup$
  • $\begingroup$ The problem doesn't say the inner product space is finite dimensional. So I'm not supposed to discuss the dimension of the orthogonal complement. $\endgroup$ – Fluffy Skye Jun 4 '18 at 2:07
  • 2
    $\begingroup$ But I think we can restrict to the space W + W' and argue the dimension. $\endgroup$ – Fluffy Skye Jun 4 '18 at 2:16
  • $\begingroup$ I solved it based on your idea. $\endgroup$ – Fluffy Skye Jun 4 '18 at 5:05
  • $\begingroup$ @FluffySkye True, I had missed that. I'll edit accordingly. $\endgroup$ – Saucy O'Path Jun 4 '18 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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