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Revision difference : Hadamard's inequality
Version 4 Version 3
Let $A=(a_{ij})$ with $1\leq i,j\leq n\in\mathbb{N}$ be a square matrix with complex coefficients. Then the following inequality holds: Let $A=(a_{ij})$ with $1\leq i,j\leq n\in\mathbb{N}$ be a square matrix with complex coefficients. Then the following inequality holds:
$$|\det(A)|\leq \prod_{i=1}^n\left(\sum_{j=1}^n|a_{ij}|^2\right)^\frac{1}{2}.$$ $$|\det(A)|\leq \prod_{i=1}^n\left(\sum_{j=1}^n|a_{ij}|^2\right)^\frac{1}{2}.$$
Moreover, if $A$ is Hermitian and positive semi-definite, the following inequality holds: Moreover, if $A$ is hermitian and positively semi-definite, the following inequality holds:
$$\det(A)\leq \prod_{i=1}^n a_{ii},$$ $$\det(A)\leq \prod_{i=1}^n a_{ii},$$
with equality if and only if $A$ is diagonal. with equality if and only if $A$ is diagonal.