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!


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.

  • $\begingroup$ 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

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