In order to define the rank of a matrix, I want to use reduced row echelon form (rref). I have an ugly proof that the rref is unique in the following sense: If $A,B$ are in rref and $A=ZB$ with invertible $Z$ then $A=B$.
However, the claim is too strong: I only need that the number of pivot positions is unique. Then the question reduces to a hunt for a proof of the following statement:
Let $F$ be any field, $Q\in GL_m(F)$, $R\in GL_n(F)$, $0\le q,r\le \min(m,n)$ with
$$
Q \pmatrix{ I_q & 0 \\ 0&0} = \pmatrix{ I_r & 0 \\ 0&0} R.
$$
Then $q=r$.
Is there a (nice) proof that only uses matrix-multiplication based arguments?
Note, that I cannot use rank or any other advanced concept here like determinant, dimension, etc.
Best Answer
Assume $q<r$. Let me partition $R=\pmatrix{ R_{11} & R_{12}\\ R_{21} & R_{22}}$ with $R_{11}\in K^{r,q}$. Then $$ \pmatrix{ I_r& 0 \\ 0&0} R= \pmatrix{ I_r& 0 \\ 0&0}\pmatrix{ R_{11} & R_{12}\\ R_{21} & R_{22}} =\pmatrix{ R_{11} & R_{12}\\ 0&0}. $$ Due to the assumption, the last $n-q$ columns are zero, so $R_{12}=0$.
In addition, there exists $x\in K^{n-q,1}$, $x\ne0$, such that $R_{22}x=0$. This can be constructed from the rref of $R_{22}$, since $R_{22}$ has more columns ($n-q$) than rows ($n-r$). Then $R\pmatrix{0\\ x}=0$, which is a contradiction to the invertibility of $R$. Hence $q\ge r$.
The inequality $q\le r$ can be obtained by multiplying the assumption by $Q^{-1}$ and $R^{-1}$ and applying the first part of the proof again.