Let $A$ and $B$ be matrices of dimensions $d \times n$. Let $C = AB^{\top}$. We also know that $C = I \mathrm{diag}(\gamma) J$ for some matrices $I$ and $J$ and vector $\gamma$ of length $m$, $m < \min(d,n)$ (i.e. $I$ is of dimension $d \times m$ and $J$ is of dimension $m \times d$). Using *only* the matrix $B^{\top}A$ (and not $AB^{\top}$, $I$, $J$ or $\gamma$), I want to find $U$ and $V$ of dimensions $m \times d$ such that $U I$ and $V J$ are invertible and $U A$ and $V B$ can be calculated. You can apply any decomposition or extract any information you need from $B^{\top} A$. FIX: I also have $B^{\top} B$ and $A^{\top} A$ at my disposal. EDIT: I managed to refine the question. Let $\sigma(D)$ be the non-zero eigenvalues of a square matrix $D$ and let $s(C)$ be the non-zero singular values of a matrix $C$. We know that if $C = AB^{\top}$, then: $s^2(C) = \sigma(C C^{\top}) = \sigma(AB^{\top} B A^{\top}) = \sigma(A^{\top} A B^{\top} B)$ I am also assuming that I can compute $A^{\top} A$ and $B^{\top} B$, which means that $s^2(C)$ is computable. I think $U$ and $V$ that I am looking for could come from the right and left singular vectors of $C$, but I don't know how to compute these singular vectors based on $A^{\top} A$, $B^{\top} B$ and $B^{\top} A$. Worst even, even if I had these $U$ and $V$, it is not clear how to compute $UA$ and $VB$ without explicitly representing $A$ and $B$. EDIT 2: I think I managed to refine the question some more: Assume we have the matrix equation: $(U^{\top} A) (Q \Lambda Q^{\top}) (A^{\top} V) = \Sigma$ such that all matrices in this equation have real values and ($d > n$): $U$ is $d \times d$ and is orthonormal $A$ is $d \times n$ $Q$ is $n \times n$ and is unitary $\Lambda$ is $n \times n$ and is diagonal (with only non-negative values) $V$ is $d \times d$ and is orthonormal $\Sigma$ is $d \times d$ and is diagonal (with only non-negative values) Is there a way to extract the matrix $U^{\top} A$ or $A V^{\top}$ if we know: $\Sigma$, $\Lambda$ and $Q$ but don't know $A$, $U$ and $V$? (I think this can be phrased more "mathematically" as a question about a uniqueness of a solution, but not sure how.) I wouldn't mind getting $U^{\top} A$ or $A V^{\top}$ up to an invertible linear transformation, as long as this transformation is a function of $U$ and $V$ only.