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I have an m*n (m>n) n-rank matrix (let's denote it by A), with nonnegative elements. SVD decomposition says, that A=UDV', where U and V are orthogonal matrixes, and their columns are the singular vectors. I would like to express UV' by only using matrix A and the singular values, but my algebra knowledge not enough. Do you have any idea?

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  • $\begingroup$ Please use MathJax for formatting: meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference $\endgroup$ – Bobson Dugnutt Mar 6 '16 at 15:34
  • $\begingroup$ You cannot do that because $UV^*$ only exists if $m = n$. Note that $U$ is $m\times m$ and $V$ is $n\times n$. Or do you use another definition of the SVD? $\endgroup$ – Friedrich Philipp Mar 6 '16 at 15:37
  • $\begingroup$ There are other definitions of the SVD, where $V$ and $D$ are square, but $U$ isn't. In this case, $UV^*$ makes sense. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 15:49
  • $\begingroup$ Sorry, I wasn't precise. The marix I'd like to compute, is the n element sum of dyadics composed of the columns of U and V. So $$ A = \sum_{i=1}^n sigma_iu_iv_i^*$$ and instead, $$\sum_{i=1}^n u_iv_i^*$$ needed. $\endgroup$ – Peter Mar 7 '16 at 8:40
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Assuming that $U$ is $m\times n$ and $V$, $D$ are $n\times n$:

You have $A = UDV^*$, thus $A^TA = VD^2V^*$. Hence $(A^TA)^{-1/2} = VD^{-1}V^*$. Hence, $A(A^TA)^{-1/2} = AVD^{-1}V^* = UDD^{-1}V^* = UV^*$. So, you have $UV^*$ expressed only in terms of $A$. Maybe this helps.

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  • $\begingroup$ Thank you for the fast answer, but the problem is the dimension of A. It is such a big matrix, that i have to monte-carlo simulate its elements....or as an alternative, conditioning the matrix could also help(I prefer diagonal scaling, or something similar due to the RAM capacity) , but I haven't succeeded yet... $\endgroup$ – Peter Mar 6 '16 at 17:43

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