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Let $A_{m\times n}$,$B_{s\times n}$ be real matrices.

Prove that if $m+s<n$, then there exists nonzero $x \in \ker A \cap \ker B$.

Could you help me, please?

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  • $\begingroup$ Welcome to Math SE! Where does this problem come from? Also, you should give a brief outline of what you've tried, so that we can help address what you do not understand. $\endgroup$ – user21820 Jul 5 '14 at 8:15
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By the Rank-nullity theorem $$ \dim\operatorname{Im}A = n - \dim \ker A\\ \dim\operatorname{Im}B = n - \dim \ker B $$ and \begin{align} \dim (\ker A \cap \ker B) &= \dim \ker A + \dim \ker B -\dim(\ker A+\ker B)\\ &\geq\dim \ker A + \dim \ker B -n\\ &= 2n-\dim\operatorname{Im}A - \dim\operatorname{Im}B-n\\ &=n-\dim\operatorname{Im}A - \dim\operatorname{Im}B\\ &\geq n-m-s>0 \end{align}

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  • $\begingroup$ How do you know, that $\dim Im A > m$ and $\dim Im B > s$? $\endgroup$ – Josef Jul 5 '14 at 8:29
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    $\begingroup$ @Josef: it is the reverse: $\dim\operatorname{Im}A\leq m$, because it is the number of independent columns, that are at most $m$. So, changing sign, $-\dim\operatorname{Im}A\geq -m$. The same holds for $B$. $\endgroup$ – enzotib Jul 5 '14 at 8:33

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