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Let $A,B$ be two $m\times n$ real matrices. Then $$|AA'|\cdot |BB'|\geq |AB'|^2.$$

For square matrices, it is the equality. How to prove this inequality then?

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    $\begingroup$ Hint: Cauchy-Schwarz inequality; especially in the case where these matrices are vectors. $\endgroup$ – Squirtle Mar 22 '14 at 0:11
  • $\begingroup$ By $ A^{'} $, do you mean transpose? $\endgroup$ – user122283 Mar 22 '14 at 0:12
  • $\begingroup$ I think you need square roots on the left-hand-side. $\endgroup$ – Martin Argerami Mar 22 '14 at 1:27
  • $\begingroup$ @Squirtle I do not see what is the norm and inner product... $\endgroup$ – XLDD Mar 22 '14 at 11:11
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Hint. If $m>n$, both sides are zero. If $m\le n$, by singular value decomposition, you may assume that $A=\pmatrix{D&0}$ and $B=\pmatrix{X&Y}$ for some diagonal matrix $D$, square matrix $X$ and $m\times(n-m)$ matrix $Y$. Using the identities $\det(PQ)=\det(P)\det(Q)$ and $\det(P^T)=\det(P)$ for square matrices, you may rewrite the alleged equality as $\det[D(XX^T+YY^T)D]\ge\det(DXX^TD)$. How do the two positive semidefinite matrices $D(XX^T+YY^T)D$ and $DXX^TD$ compare?

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