Counterexample of Converse of "$\operatorname{rank} (PA) = \operatorname{rank} (A)$ if $P$ is invertible"

Question

I'm studying linear algebra, and got to a theorem:

Let $A$ be an $m \times n$ matrix. If $P$ and $Q$ are invertible $m \times m$ and $n \times n$ matrices, respectively, Then

(a) $\operatorname{rank} (AQ) = \operatorname{rank} (A)$

(b) $\operatorname{rank} (PA) = \operatorname{rank} (A)$

I know how to prove these two, but I want to know counterexamples for the converse of these two respectively.

The converse proposition doesn't make sense, right? Then, for (a), I need a linear transformation $T_A$ and $T_Q$ whose $\operatorname{rank} (AQ) = \operatorname{rank} (A)$ but $T_Q$ is not bijective and $Q$ is not invertible. How can I construct it, then?

Answer 1

Let $A = \left(\begin{array}{cc} 1&0\\0&0\end{array}\right)$. Then $\textbf{rank}(AA) = \textbf{rank}(A)$ because $A^2 = A$, and yet $A$ is not invertible.