$\operatorname{rank}AB\leq \operatorname{rank}A, \operatorname{rank}B$

Question

Prove that if $A,B$ are any such matrices such that $AB$ exists, then $\operatorname{rank}AB \leq \operatorname{rank}A,\operatorname{rank}B$.

I came across this exercise while doing problems in my textbook, but am not sure where to start for the proof of this. I think columnspace might be involved in the proof, although I am not sure.

Rank is the amount of leading 1s in Reduced Row Echelon Form of a matrix. — Sujaan Kunalan, Jun 14 '13 at 03:06
Oh, I didn't realize "rankA, rankB" was synonymous with "min(rank(A),rank(B)). I thought the question was saying that rank(AB) was less than or equal to both rank A, and rank B. I must've misread. — Sujaan Kunalan, Jun 14 '13 at 04:57

score 0 · Accepted Answer · answered Jun 14 '13 at 03:09

I think it is best to prove two things that are stronger statements:

$x\in \ker B \Rightarrow x\in \ker AB$,

$y\in \text{im}\;AB \Rightarrow y\in \text{im}\;A$.

Together with the rank-nullity theorem, I believe this provides a solution not in the link @Amzoti provided above. Edit- nevermind, it's in there, but not all in one place.

$\operatorname{rank}AB\leq \operatorname{rank}A, \operatorname{rank}B$

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