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I have to show that

If $A$ and $B$ be two matrices of the same order $n$, then $$ \text{rank}\, A + \text{rank}\,B \le\text{rank}\,AB + n $$

In a previous exercise I have already shown that

$$ \text{dim}\,(\text{ker}\,(AB)) \le \text{dim}\,(\text{ker}\,(A)) + \text{dim}\,(\text{ker}\,(B)) $$

Is it enough to use the rank-nullity theorem and subbing in

$$ \text{dim}\,(\text{ker}\,(AB))= \text{dim}\, (V) - \text{rank}\,(AB) $$

to show that the two inequalities are basically saying the same thing?

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    Of course! It's just another form of what you've already shown. – 2017-01-20
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    http://math.stackexchange.com/questions/298836/sylvester-rank-inequality?rq=1 – 2017-01-20
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    http://www.m-hikari.com/imf-password2009/33-36-2009/luIMF33-36-2009.pdf – 2017-01-20

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