# To what extent are the factors in a matrix equivalence unique?

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.

• It might be helpful to note that this is equivalent to finding a rank-factorization of $Q$ Aug 12 '20 at 20:58

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.