How to find a solution of $XA=AM$ s. t. $X$ is invertible.

Question

Let $A$ be a complex $n\times m$ matrix ($n<m$), which is row full rank. For any $m\times m$ matrix $M$, such that $AM$ is still row full rank, can we find an invertible matrix $X$, such that $XA=AM$?

Answer 1

Hint: $A = \begin{pmatrix} 1 & 0 \end{pmatrix}$, and $M = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$.

Answer 2

Some basic facts that can be used to reason about the problem:

• If $X$ is invertible, the rowspace of $XA$ is the rowspace of $A$.
• Every row of $AM$ is in the row space of $M$.