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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?

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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.

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  • $\begingroup$ what about A is a m x n matrix case, i mean when A is not equal to Q or P? $\endgroup$ – MOON May 24 '16 at 12:31
  • $\begingroup$ @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 $\textbf{rank}(PA) = \textbf{rank}(A)$ $\endgroup$ – Alex G. May 24 '16 at 12:36
  • $\begingroup$ @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. $\endgroup$ – Alex G. May 24 '16 at 12:40

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