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If A is matrix m by n and B is matrix n by r

The rank of matrix AB ought to be minimum of rank(A) or rank(B) how may i prove this? Thanks

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Pretty sure the conclusion is false. Why do you think it's true? – Ben Millwood Sep 7 '12 at 0:15
Hint: two nonzero matrices with a product that is zero will suffice as a counterexample – Ben Millwood Sep 7 '12 at 0:16
rank($AB$) $\leq$ min(rank($A$),rank($B$)). Equality is not guaranteed. – svenkatr Sep 7 '12 at 0:18
On the other side, $\text{rank}(AB) \ge \text{rank}(A) + \text{rank}(B) - n$. – Robert Israel Sep 7 '12 at 2:46

No. The product of two rank $1$ matrices may have rank $0$. $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

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