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Suppose we have $n\times m$ matrix $Q$, and that $\exists$ $ n\times n$ matrices $P,P'$, and $m \times m$ matrices $R,R'$ all invertible, such that $PQR = \begin{bmatrix} I_{r\times r} & 0 \\ 0 & 0\end{bmatrix}= P'QR'$. Where $r$ is the rank of the matrix $Q$.

What is the relationship between $P,P',R,R'$? I know this question is not specific enough, but I'm not sure how to frame what I'm looking for properly. I want to know to what extent we can consider the factors in this decomposition unique.

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    $\begingroup$ It might be helpful to note that this is equivalent to finding a rank-factorization of $Q$ $\endgroup$ Aug 12 '20 at 20:58
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First of all, it is helpful to note that this is equivalent to considering a rank-factorization of $Q$. In particular, rewrite this in the form $$ Q = A\pmatrix{I_r & 0\\0 & 0}B $$ (in particular, we can take $A = P^{-1}$ and $B = Q^{-1}$).

if we partition $A$ into two block-columns and $B$ into two block-rows, then we have $$ Q = A\pmatrix{I_r & 0\\0 &0}B \iff Q = \pmatrix{A_1 & A_2}\pmatrix{I_r & 0\\0 &0} \pmatrix{B_1 \\ B_2}\\ \iff Q = A_1 I_r B_1 = A_1 B_1, $$ where we note that $A_1$ has linearly independent column, and $B_1$ has linearly independent rows. Once $A_1$ and $B_1$ are chosen, we need only complete the columns of $A_1$ and the rows of $B_1$ to form bases of $\Bbb R^m$ and $\Bbb R^n$.

Now, the question becomes to what extent are the choice of $F = A_1$ (size $m \times r$) and $G = B_1$ (size $r \times n$) unique in the factorization $Q = FG$? We see that for any invertible size $r$ matrix $S$, taking $F' = FS$ and $G' = S^{-1}G$ gives us a new factorization. In fact, I claim that every rank factorization can be attained in this fashion. Indeed, this can be seen as a consequence of the uniqueness of the "skinny" QR factorization.

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