# A determinant inequality

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?

• Hint: Cauchy-Schwarz inequality; especially in the case where these matrices are vectors. – Squirtle Mar 22 '14 at 0:11
• By $A^{'}$, do you mean transpose? – user122283 Mar 22 '14 at 0:12
• I think you need square roots on the left-hand-side. – Martin Argerami Mar 22 '14 at 1:27
• @Squirtle I do not see what is the norm and inner product... – XLDD Mar 22 '14 at 11:11

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?