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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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    @Rahul, yes, I guess it should be $\Sigma^+$.2012-08-01

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