# sum of singular vector dyadics derived from the matrix itself

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?

• Please use MathJax for formatting: meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Bobson Dugnutt Mar 6 '16 at 15:34
• 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? – Friedrich Philipp Mar 6 '16 at 15:37
• There are other definitions of the SVD, where $V$ and $D$ are square, but $U$ isn't. In this case, $UV^*$ makes sense. – Friedrich Philipp Mar 6 '16 at 15:49
• 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. – Peter Mar 7 '16 at 8:40

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.