Existent of matrix sequence such that $AX_nB$ is full rank

Question

Suppose I have two matrices $A\in \mathbb{R}^{p\times m}$, $B\in \mathbb{R}^{n\times r}$, where $m\geq n\geq p\geq r$ and both $A,B$ have full rank. Given a low-rank matrix $X\in \mathbb{R}^{m\times n}$. whether it exists a sequence of matrix $X_n\rightarrow X$ such that $AX_nB$ is full rank for all $n=1,2,3...$?

Thank you in advance.

Answer 1

The answer is yes. Here is an outline of one approach to proving that this is the case.

• First, note that there exists a matrix $M$ such that $AMB$ has full rank. If you are familiar with the MP pseudoinverse, then one convenient choice for such a matrix is $M = A^+B^+$.
• Show that for $t \in \Bbb R$, $X + tM$ will have low rank (i.e. will fail to have its full rank of $r$) for at most finitely many values of $t$. One helpful fact here is that a matrix $P$ of size $p \times r$ with $p \geq r$ will have low rank if and only if $\det(PP^\top) = 0$. Notably, $f(t) = \det([X + tM][X + tM]^\top)$ is a polynomial (of degree at most $2r$). You might find it helpful to consider the function $t^{2r}f(t^{-1}) = \det([tX + M][tX + M]^\top)$.
• From the previous result, we can conclude that there exists an integer $m\geq 0$ such that for all $n = 1,2,3,\dots,$ $X + \left(\frac 1{m+n} \right)M$ has full rank.