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Hi: I'm new to this list so apologies if I do anything wrong with my question.

Suppose I have a matrix $Y$ whose SVD can be decomposed as $Y = U_{0}\Sigma_{0}V_{0}^{*} + U_{1}\Sigma_{1} V_{1}^{*}$ where $U_{0}$ and $V_{0}$ are the singular vectors associated with the singular values greater than $\tau$ and
$U_{1}$ and $V_{1}$ are the singular vectors associated with singular values greater than $\tau$.

The paper then goes on to say that $U_{0}^{*} U_{1}\Sigma_{1} V_{1}^{*} = 0$ and $U_{1}\Sigma_{1} V_{1}^{*}V_{0} = 0$.

Could someone explain why the last two claims are true ?

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As far as I understand this just the orthogonality of vectors U_1 U_2, and respectively for V. – Alexander Chervov Nov 16 '12 at 21:21
reformatted and retagged – Yemon Choi Nov 16 '12 at 21:24
@mark: this question will probably elicit a more detailed response at (actually, Alexander's comment is already the answer; and in your question, you probably meant "greater than tau" and "less than tau", rather than greater than both times). – Suvrit Nov 16 '12 at 22:17
thanks Alexander and Suvrit. I have to look at a real example of an SVD to see why it can decomposed that way better ( I'll look in one of my matrix algebra books ) but I think I get it. Thanks a lot. – mark leeds Nov 17 '12 at 6:51
alexander and suvit: I was going to check out but the answer would be the same thing as Alexander's. I think my confusion is how the SVD can be broken into pieces like that. I was looking at searle's text and he sort of explains how the svd decomposition works in terms of the how one can break it into two pieces. but what is the setup so one can write it that way. By this I mean is $U = [U_{0}, U_{1}]$ and $ V = [V_{0} V_{1}]$. so that the two pieces make up all the columns of each matrix ? Thanks. Suvrit, my future questions will go to site you mentioned – mark leeds Nov 18 '12 at 1:46

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