Someone posted here that when $A$ has more rows than $B$, $\det(AB)=0$. How?

  • 4
    $\begingroup$ Because $\text{rank}(AB)\leq \min\left(\text{rank}(A),\text{rank}(B)\right)$. $\endgroup$ – Git Gud May 10 '15 at 0:21
  • $\begingroup$ I don't see where is the problem since $AB$ is a square matrix, everything is ok. $\endgroup$ – MonkeyKing May 10 '15 at 0:30
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    $\begingroup$ The intuition is that B maps from a high dimensional space to a low dimensional space, then A maps back up to the higher dimensional space. Moving through the lower dimensional space necessarily loses information, since many point are mapped to the same values, so the overall transformation cannot be inverted. $\endgroup$ – G. H. Faust May 10 '15 at 1:24

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