Decomposition of a rank-deficient matrix

Question

Given a complex or real $n\times m$ matrix $M$ with rank $r$, one can write it as $$M=LR$$ where $L$ is a $n\times r$ matrix, and $R$ is a $r \times m$ matrix.

1. Does this hold for arbitrary fields? (I'd assume so)

2. What about associative division algebras? E.g., does this hold for matrices with quaternionic entries? (If there is any well-defined notion of rank in this case.)

Answer 1

1) Yes (proof sketch below).

2) I don't know. You might want to ask a separate question.

Proof sketch for 1):

Using Gaussian elimination, you can first decompose $M = PDQ$, where D is an $n \times m$ matrix with 1s on the first $r$ diagonal elements, and zeroes everywhere else, and $P$ and $Q$ are nonsingular square matrices.

Denoting $P_r$ the $n \times r$ matrix consisting of the first $r$ columns of $P$, and $Q_r$ the $r \times m$ matrix consisting of the first $r$ rows of $Q$, you can then show using block matrix multiplication that $M = P_r I_r Q_r = P_r Q_r$.