I've found two versions of Hadamard's inequality :

(1) If $$P$$ is a $$n\times n$$ positive-semidefinite matrix, then : $$\det(P)\le\prod_{i=1}^n p_{ii} .$$ (2) For any $$M$$ is a $$n\times n$$ matrix, then : $$|\det(M)|\le\prod_{i=1}^n ||m_i || =\prod_{i=1}^n \left(\sum_{j=1}^n |a_{ij}|^2\right)^{1/2}.$$ It's easy to prove that $$(2)\Rightarrow (1)$$, but do we have $$(1)\Rightarrow (2)$$ ?

• Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix. – mouthetics Jan 3 at 16:02

From (1) we know that $$\sqrt{\det(MM^T)}\le\prod_{i=1}^n\{MM^T\}_{ii}^{1/2}=\prod_{i=1}^n \left(\sum_{j=1}^n |m_{ij}|^2\right)^{1/2}=\prod_{i=1}^n ||m_i ||$$
Since $$\det(MM^T)=\det(M)\det(M^T)=\det(M)^2$$
we have $$|\det(M)|=\sqrt{\det(MM^T)}\le\prod_{i=1}^n ||m_i ||$$.