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)


  • $\begingroup$ 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

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 $V$.


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