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Let $A \in \mathbb{R}^{d \times n}$ and let $B \in \mathbb{R}^{d \times n}$ where $d > n$.

Let $C = A \times B^{\top}$.

Let $U \Sigma V^{\top}$ be the SVD of $C$.

Can we say anything special about the matrix:

$$A A^{\top} (U \Sigma^{-1} V^{\top}) B$$ ?

I am especially interested in finding out about the sum of the values in this matrix.

Edit: I am pretty sure that $A^{\top} (U \Sigma^{-1} V^{\top}) B = I$. (I mocked around with that in Matlab. No real proof - but I also have a "feeling" that a proof would involve the pseudo-inverse of $AB^{\top}$ because $V \Sigma^{-1} U ^{\top}$ is the pseudo-inverse of $AB^{\top}$.

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    If you replace $A$ with $2A$, your matrix also doubles, so the sum of its values can't always be $n$. Maybe the $AA^T$ in the beginning should be just $A^T$?2012-07-31
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    perhaps it doesn't sum up exactly to $n$, but there is something about this matrix and the sum of its elements.2012-07-31
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    I changed the question following your comment.2012-07-31
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    By $\Sigma^{-1}$ you mean the *pseudo*-inverse, right? $\Sigma$ is an $d\times d$ matrix with rank at most $n < d$. With the pseudoinverse your equation $A^T(U\Sigma^+V^T)B=I$ does appear to be true, but I haven't been able to see a way to prove it.2012-07-31
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    Actually, it looks like it's only true if $A$ and $B$ have full rank...2012-07-31
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    @Rahul, yes, I guess it should be $\Sigma^+$.2012-08-01

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