If a matrix $B$ has full rank by columns, can we conclude that $rg (AB) = rg (A)$? And what $rg (BA) = rg (A)$? The matrices $A$ and $B$ are not necessarily square, in addition it must be taken into account that full rank by columns means that the range of $B$ is equal to its number of columns.
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$\begingroup$I am assuming you are talking about the image. Let's assume that $A$ and $B$ are $n$-dimensional, then yes $rg(AB) = rg(A)$. This comes from the fact that you have $Ax= ABy,$ where $y = B^{-1}x.$
The other equality is wrong. Take
$$ B = \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right) $$
$$A = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) $$
Then,
$$BA = \left( \begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right) $$
Clearly, $rg(BA) \neq rg(A)$
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