I don't know if I'm missing something out but is this just not proved by saying ${\rm Im} (AB)$ subspace of ${\rm Im}(A)$ so ${\rm rank}(AB) \leq {\rm rank}(A)$?
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1It is, except you should replace in the proof $A$ and $B$ with the associated linear maps (after a basis has been chosen). – Bernard May 18 '17 at 23:51
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Take a look at this post, this question was already asked. – BB3C May 18 '17 at 23:52
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@Bernard Would this result be different if, say, $A$ and $B$ were not $n \times n$ matrices (but still defined such that their product made sense)? – PhysicsMathsLove May 18 '17 at 23:52
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@BBC3 Okay, but what is it about these being $n \times n$ that allows you to say it is specifically less than $rank(A)$ rather than $\text{min} { rank(A), rank(B) }$ – PhysicsMathsLove May 18 '17 at 23:54
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It would be exactly the same. – Bernard May 18 '17 at 23:55
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BBC3 cannot reply to your comment because s/he doesn't have enough reputation to leave comments. His/her comment above was posted as an answer then converted into a comment. See https://math.meta.stackexchange.com/q/26391/18398 – JRN May 19 '17 at 00:49
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1@PhysicsMathsLove, by definition of "minimum," $\min(a,b)\le a$ no matter what the setting is. – Barry Cipra May 19 '17 at 00:59
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Got it now - cheers! – PhysicsMathsLove May 19 '17 at 01:11