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Let $0<a<b$, 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.

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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: $$ \pmatrix{A^{1/2}\\&B^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&B^{1/2}} = \pmatrix{0&A^{-1/2}B^{1/2}\\B^{-1/2}A^{1/2} & 0}\\ % \pmatrix{A^{1/2}\\&A^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&A^{1/2}} = \pmatrix{0&I\\A^{1/2}B^{-1}A^{1/2} & 0} $$

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  • $\begingroup$ 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 $B$. $\endgroup$
    – user344045
    Commented Nov 19, 2019 at 19:59

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