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If $B$ is invertible, show that rank($AB$) = rank($A$).
I've seen this question asked elsewhere but all had answers I didn't understand. I know how to solve the following problem
If $A$ is invertible, then rank($AB$) = rank($B$)
Because if $Bx=0$, then $ABx = A0 = 0$, and when $ABx=0$ then $Bx=0$ because $A$ is invertible, so null($AB$)=null($A$), and by the rank-nullity theorem, rank($A$) = rank($AB$).
However when $B$ is invertible, as in the problem we have to tackle, I don't know how to use that fact. $ABx = 0$, but $B$ is in the middle so we can't simply get rid of it to get a meaningful expression.
Does someone know how to tackle this?