Suppose $A\in M_{n\times n}(\mathbb{C})$, $H = (A + A^*)/2$, and $H$ is positive definite. Prove that $\det H\leq |\det A|$.
Positive definite and Determinant
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linear-algebra
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0It would improve your Question to add some context. Where did you find this problem? What related facts do you think might come into the proof? Or even, what makes this problem interesting to you? – 2017-01-08
1 Answers
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I will help to get you started.
Begin by writing $A = H + S$, where $S = (A - A^*)/2$ is the skew-Hermitian part of $A$. Since $H$ is invertible it follows that $A = H(I + H^{-1}S)$ and so $$ |\det(A)| = \det(H)|\det(I + H^{-1}S)| $$ So your goal is to show that $|\det(I + H^{-1}S)| \geq 1$. Here is a hint: Prove that $H^{-1}S$ is similar to a skew-Hermitian matrix. The real, symmetric positive definite square root of $H$ may be useful.