When does $\det(\frac{A+A^T}{2})=\det(A)$ for positive-definite $\frac{A+A^T}{2}$?

Question

Setup: Let $A$ be a real square matrix and assume its symmetric part $\frac{A+A^T}{2}$ is positive-definite. The inequality $$ \det\left(\frac{A+A^T}{2}\right) \leq \lvert\det(A)\rvert $$ is known as in this mathoverflow comment. It is a consequence of the Fan–Hoffman inequality which states that $\lambda_i(\frac{A+A^T}{2})\leq \sigma_i(A)$ where $\sigma_i$ is the $i$th singular value.

Question: I am interested in knowing when we have equality $\det\left(\frac{A+A^T}{2}\right) = \det(A)$. I believe this equality is achieved if and only if $\frac{A+A^T}{2}=A=A^T$. I have a somewhat long proof of this claim, but I was wondering if there is a short reason or a reference proving this fact.

Answer 1

Write $A = B + C$ where $B$ is symmetric and $C$ is antisymmetric. The assumption is that $B$ is positive-definite and $\det(A) = \det(B)$. we wish to prove $C=0$.

First, every symmetric positive-definite matrix is of the form $g \cdot g^t$ for some $g \in \mathrm{GL}_n(\mathbf{R})$. Thus, by acting by this group, we can assume $B =1$. In more detail, if $B = g \cdot g^t$, take the equation $A = B + C$ and multiply on the left by $g^{-1}$ and on the right by $g^{-t}$. Let $A_1 = g^{-1} A g^{-t}$ and $C_1 = g^{-1} C g^{-t}$. Then $A_1 = 1 + C_1$. Observe that $C_1$ is still anti-symmetric. Moreover, because $\det(A) = \det(B)$, $\det(A_1) = 1$.

Now, as $C_1$ is anti-symmetric, by the spectral theorem I can assume $C_1$ is block diagonal with two-by-two anti-symmetric blocks. (See the Wikipedia page for "Skew-symmetric matrix"). More specifically, $C_1 = h C_2 h^{t}$ for some $h \in O(n)$ and $C_2$ block-diagonal with anti-symmetric $2 \times 2$ blocks. Write $A_2 = h^{-1} A_1 h^{-t}$. We have $A_2 = 1 + C_2$, and $\det(A_2) = 1$.

The determinant of $1 + C_2$ is now immediate to compute, and one gets $C_2 = 0$ from the equality $1 = \det(1+C_2)$. As $C_2 = 0$, $C_1 = 0$ and thus $C=0$.

Answer 2

As said in the question, with $S=(A+A^T)/2$, multiplying the inequalities $0<\lambda_i(S)\le\sigma_i(A)$ give the inequality $0 < \det S \le \det A$.

Now $\det A= \det S$ implies that each inequality $\lambda_i(S)\le\sigma_i(A)$ is an equality for each $i$, as a single strict inequality would lead to a strict inequality for the determinants. Hence the Frobenius norms of $S$ and $A$ are equal, so $$0=4\|S\|^2-4\|A\|^2 =2\langle A,A^T\rangle-2\|A\|^2. $$ This gives $ \|A^T-A\|^2 =0$ by expanding the square.

Answer 3

As $A$ is real, and normal, it commutes with $A^\top$, and these matrices can be simultaneously diagonalised by a unitary $U$. That is, both sides of $\det((A+A^\top)/2)=\det A$ can be conjugated by $U$ to obtain $\det(D/2+D'/2)=\det D$, with $D,D'$ diagonal matrices, and multisets of their entries are equal, each entry a positive real.

In the minimal nontrivial case of 2×2 matrices, assuming $A\neq A^\top$, the latter equality becomes $D_1 D_2=(D_1/2+D_2/2)^2$, with $D_2<D_1$, and $D=\operatorname{diag}(D_1,D_2)$, $D'=\operatorname{diag}(D_2, D_1)$. But this is only possible if $D_1=D_2$, a contradiction showing that $A=A^\top$ in this case. In general, it's not immediately clear how to reach the same conclusion.