Suppose $B$ has a full column rank. Does it hold that $\operatorname{rank}(AB)=\operatorname{rank}(A)$?
I found a similar post. But it does not prove or disprove the above statement.
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The statement is only true for $B$ square otherwise, as counterexample, we can consider $A_{n\times n} $ full rank and $B_{n\times 1} $ such that rank$(AB)=1$
$$A=\begin{pmatrix} 1&1 \\1&-1 \end{pmatrix}, B=\begin{pmatrix} 1\\0 \end{pmatrix} \implies AB=\begin{pmatrix} 1\\1 \end{pmatrix}$$
which leads to rank$(AB)=1$ with rank$(A)=2$.
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why $Av$ does not equal to 0? – mxdxzxyjzx Nov 16 '19 at 09:10
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@mxdxzxyjzx I've revised with a simpler way to see that, indeed the first one wasn't completely correct and enlightening. – user Nov 16 '19 at 09:37