Let $0<a<b$, and let $A, B \in \mathbb{R}^{n \times n}$ be two positive definite symmetric matrices.
My question is: is the matrix $M \in \mathbb{R}^{2n \times 2n}$ be defined by \begin{align*} M = \begin{bmatrix} bA & aB\\ aA & bB \end{bmatrix} \end{align*} positive definite or not? (In the sense $x^\intercal (M + M^\intercal) x >0$ for any nonzero $x \in \mathbb{R}^{2n}$.)
I tried the following.
We know that the symmetric matrix $P$ defined by \begin{align*} P = \begin{bmatrix} bA & a(AB)^{\frac{1}{2}}\\ a(BA)^{\frac{1}{2}} & bB \end{bmatrix} \end{align*} is positive definite, because it writes as \begin{align*} P = L^\intercal D L, \qquad \text{for }D = \frac{1}{2} \begin{bmatrix} (a+b)A & 0 \\ 0 & (b-a)A \end{bmatrix}, \quad L = \begin{bmatrix} I_n & A^{-\frac{1}{2}} B^{\frac{1}{2}}\\ I_n & - A^{-\frac{1}{2}} B^{\frac{1}{2}} \end{bmatrix} \end{align*} and $D$ is positive definite (hence, $x^\intercal P x = (Lx)^\intercal P (Lx)>0$ for any nonzero $x \in \mathbb{R}^{2n}$).
Moreover, one can see the following relationship between $M$ and $P$: \begin{align*} M = C^{-1} P C, \qquad \text{for } C = \mathrm{diag}(A^\frac{1}{2}, B^{\frac{1}{2}}) \end{align*} Hence, we know that $M$ and $P$ have the same eigenvalues. However, $M$ is not symmetric, so I don't know if this information is of any use.
It is not always positive definite. Let $C=A^{-1/2}BA^{-1/2}$. Since $$ \pmatrix{A^{-1/2}\\ &A^{-1/2}}(M+M^T)\pmatrix{A^{-1/2}\\ &A^{-1/2}}=\pmatrix{2bI&a(I+C)\\ a(I+C)&2bC} $$ and $C$ can be unitary diagonalised, $M+M^T$ is positive definite if and only if $\pmatrix{2b&a(1+c)\\ a(1+c)&2bc}$ is positive definite for every eigenvalue $c$ of $C$. By Sylvester's criterion, the latter condition is equivalent to $4b^2c-a^2(1+c)^2>0$, which clearly isn't always satisfied.