I have a matrix $A$, and I can run svd on it to get $UDV'$. Of course, the $U$ and $V$ are not unique, and I'm looking for a particular pair of $U$ and $V$. I want to find $U$ and $V$ such that the last $k$ rows of $U$ (call it $U_2$) and the last $k$ columns of $V'$ (call it $V_2'$) are equal. If it helps, I'm reasonably convinced that the diagonal $U_2*D*V_2'$ would be all 1s and possibly $0$s on the off-diagonals.
2026-03-29 16:54:29.1774803269
Constrained SVD
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If $A$ is not the identity matrix it's obviously impossible to have $U_2 D V_2' = A$ as the identity Matrix. Or have I got something wrong in your definition of "off-diagonals"?