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Let $A$ be an $n \times n$ matrix with real entries and let $H = \frac{A+A^T}{2}$. Assume that $H$ is positive definite. Prove that $\det(H) \geq \det(A)$.

This question is obtained from Moscow (I don't have the specific source).

My attempt: Note that $H$ is a symmetric matrix with real entries and positive definite. Hence, it is diagonalizable with positive eigenvalues, i.e. $H = P^T D P$ where $P^T \cdot P=1$ and $\det(H) = \prod \lambda$ where $\lambda$ is an eigenvalue of $H$.

I don't know how to relate eigenvalues of $H$ to eigenvalues of $A$. Can anyone give some hints?

Idonknow
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2 Answers2

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Claim: $\det(A^{-1}A^\top+I)\geq 2^n$.

Proof: Note that $\det(A^{-1}A^\top)=1$. Let $\lambda_1, \cdots, \lambda_n$ be the eigenvalues of $A^{-1}A^\top$. Then $\lambda_1\cdots\lambda_n=1$ and the eigenvalues of $A^{-1}A^\top+I$ are $\lambda_1+1, \cdots, \lambda_n+1$. So \begin{align*} \det(A^{-1}A^\top+I)&=(\lambda_1+1)\cdots(\lambda_n+1)\\ &\geq (2\sqrt{\lambda_1})\cdots(2\sqrt{\lambda_n})\\ &=2^n \end{align*}

Now we have $\displaystyle\det H=\det\left(\frac{A+A^\top}{2}\right)=\det(A)\det\left(\frac{I+A^{-1}A^\top}{2}\right)\geq\det(A)$. It is easy to see that equality holds if and only if $A^{-1}A^\top=I$, i.e. $A$ is symmetric (under the assumption that the eigenvalues of $A$ are positive).

Alex Fok
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I guess the basic field you concerned is $\mathbb R$? Then by the hypothesis, we can write $A=H+S$, where $H$ is a positive definite symmetric matrix and $S$ is skew-symmetric. Then since $H$ is positive definite, there is a invertible matrix $P$ such that $P^THP=I$, note that $P^TSP$ is still skew-symmetric, so there is a orthogonal matrix $O$ such that $O^TP^TSPO$ is of canonical form, then to compare $\det(A)$ with $\det(H)$ is equivalent to compare $\det(O^TP^TAPO)$ with $\det(O^TP^THPO)$ (note that $O^TP^TAPO=O^TP^THPO+O^TP^TSPO=I+O^TP^TAPO$, and $O^TP^TAPO$ is of canonical form).

Censi LI
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