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

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?

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.
• @MOON Take $A$ to be the $n\times 2$ matrix (for any $n\geq 1$) with a $1$ in the upper left-hand corner and $0$'s elsewhere. Let $Q$ be the same $2\times 2$ matrix as in my answer. Then $AQ = A$, so $\textbf{rank}(AQ) = \textbf{rank}(A)$. Likewise for (b), let $A$ be the $2\times n$ matrix with a $1$ in the upper left-hand corner and $0$'s elsewhere and let $P=Q$. Then $PA=A$ so $\textbf{rank}(PA) = \textbf{rank}(A)$ – Alex G. May 24 '16 at 12:36
• @MOON More generally still, $A$ could be taken to be the $m\times n$ matrix with a $1$ in the upper left-hand corner and $0$'s elsewhere. Then $P$ can be taken to be the $m\times m$ matrix also with only a $1$ in the upper left-hand corner, and $Q$ can be the same $n\times n$ matrix. Then $PA = A$ and $AQ = Q$, so we have very general counterexamples now. – Alex G. May 24 '16 at 12:40