1
$\begingroup$

So, I was reading a paper, and a step in the proof stumbled me.

$$ U \in \mathbb{R}^{m \times k} (GIVEN: Orthogonal \ matrix), M \in \mathbb{R}^{m \times n}, M = U^*\Sigma^*{V^*}^T (rank-k \ SVD) $$

$$ ||(I - UU^T)U^*\Sigma^*{V^*}^T||^2_2 = || U_{perp}U^*\Sigma^*||^2_2 $$ where $U_{perp}$ is the basis of the orthogonal complement of the subspace spanned by the columns of U.

I have toiled hard to understand this, but I cannot see why this should hold.

Reference: https://arxiv.org/pdf/1212.0467v1.pdf. Lemma: C.1 (Page - 34)

Thanks.

$\endgroup$
2
  • $\begingroup$ I suppose you mean "toyed" and not "toiled" $\endgroup$
    – b00n heT
    Sep 4 '16 at 10:24
  • $\begingroup$ I consider myself not so good with linear algebra. So, in my humble capability, I tried hard to understand the identity but could not. I would be grateful to you if you can help an amateur in this regard. $\endgroup$ Sep 4 '16 at 11:19
1
$\begingroup$

It may help to note that $UU^T$ is the projection onto the span of the columns of $U$, and that $I-UU^T$ is the projection onto the orthogonal complement. We can therefore write $$ I-UU^T= U_{perp}U_{perp}^T $$ I think that should help. Not also that $\|A\|=\|AU\|=\|VA\|$ for any orthogonal matrices $U$ and $V$.

$\endgroup$

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.