# Find the right positive definite block matrix

Let $$0, and let $$A, B \in \mathbb{R}^{n \times n}$$ be two positive definite symmetric matrices (that do not commute).

My question is: are the matrices $$M, P \in \mathbb{R}^{2n \times 2n}$$ defined by \begin{align*} M = \begin{bmatrix} bABA & aAB\\ aBA & bBAB \end{bmatrix},\qquad P = \begin{bmatrix} bA & aAB\\ aBA & bB \end{bmatrix} \end{align*} positive definite or not? (In the sense $$x^\intercal M x >0$$ for any nonzero $$x \in \mathbb{R}^{2n}$$.)

Why this question: I am trying to find a positive definite symmetric matrix $$Q \in \mathbb{R}^{2n \times 2n}$$ such that the product \begin{align*} Q \begin{bmatrix} 0 & A^{-1}\\ B^{-1} & 0 \end{bmatrix} \end{align*} is a symmetric matrix. My candidates at the moment are M and P, which are such that \begin{align*} M \begin{bmatrix} 0 & A^{-1}\\ B^{-1} & 0 \end{bmatrix} = \begin{bmatrix} aA & bAB\\ bBA & aB \end{bmatrix}, \qquad P \begin{bmatrix} 0 & A^{-1}\\ B^{-1} & 0 \end{bmatrix} = \begin{bmatrix} aA & bI_n\\ bI_n & aB \end{bmatrix} \end{align*} but I don't know if they are positive definite.

What I tried: I tried to find an invertible matrix $$K$$ such that the product $$K^\intercal P K$$ (or $$K^\intercal M K$$ ) is easier to study, since $$P$$ is positive definite if and only if $$K^\intercal P K$$ is. For example, for \begin{align*} K = \begin{bmatrix} A^{-1} & A^{-1}\\ B^{-1} & B^{-1} \end{bmatrix} \end{align*} we have \begin{align*} K^\intercal P K = \begin{bmatrix} bA^{-1} + 2a I_n + bB^{-1} & bA^{-1} - bB^{-1} \\ bA^{-1} - bB^{-1} & bA^{-1} - 2a I_n + bB^{-1} \end{bmatrix}. \end{align*} But I don't know how to proceed for this matrix either.

First of all, we must assume that $$b>0$$ for either matrix to be positive definite.
For $$M$$, we compute the Schur complement $$bABA - \frac{a^2}{b}(AB)(BAB)^{-1}(BA)= bABA - \frac{a^2}{b}A.$$ Since $$bABA$$ is positive definite, $$M$$ will be positive definite if and only if $$bABA - \frac{a^2}{b}A$$ is positive definite. With the Loewner order, we can write this condition as $$bABA - \frac{a^2}{b}A > 0 \iff bABA > \frac{a^2}{b}A \iff \\ ABA > \frac{a^2}{b^2} A \iff A^{-1/2}(ABA)A^{-1/2} > A^{-1/2}[\frac{a^2}{b^2} A]A^{-1/2} \iff\\ A^{1/2}BA^{1/2} > \frac{a^2}{b^2}I.$$ That is, $$M$$ will be positive definite if and only if the eigenvalues of $$A^{1/2}BA^{1/2}$$ are greater than $$a^2/b^2$$. We see that the matrix $$AB$$ is similar since $$AB = A^{1/2}(A^{1/2}BA^{1/2})A^{-1/2},$$ so $$M$$ will be positive definite if and only if the eigenvalues of $$AB$$ are greater than $$a^2/b^2$$.
For $$P$$, we compute the Schur complement to be $$bA - \frac{a^2}{b} ABA$$ and a similar analysis can be applied. We find that $$P$$ will be positive definite if and only if either $$a=0$$ or the eigenvalues of $$AB$$ are less than $$b^2/a^2$$.
In summary, $$M$$ and $$P$$ will be positive definite if and only if $$a=0$$ or the eigenvalues of $$AB$$ lie inside the interval $$(a^2/b^2,b^2/a^2)$$.
A potentially helpful insight: we can reduce your original problem to a possibly simpler case by considering a congruent matrix. For instance: $$\pmatrix{A^{1/2}\\&B^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&B^{1/2}} = \pmatrix{0&A^{-1/2}B^{1/2}\\B^{-1/2}A^{1/2} & 0}\\ % \pmatrix{A^{1/2}\\&A^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&A^{1/2}} = \pmatrix{0&I\\A^{1/2}B^{-1}A^{1/2} & 0}$$
• Thank you for your answer and the the insight. I will now try to find a $Q$ that doesn't put such constraints on $A$ and $B$. Commented Nov 19, 2019 at 19:59