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Suppose we have a positive definite matrix $Q \in \mathbb{R}^{n\times n}$, are there general conditions on $A \in \mathbb{R}^{m\times n}$ such that $A Q^{-1}A^T$ is positive semi-definite ($A Q^{-1}A^T \succeq 0$)? Or positive definite ($A Q^{-1}A^T \succ 0$)?

Extension: Suppose there are no general conditions on $A$ such that $A Q^{-1}A^T$ can be made positive semi-definite or positive definite given that $Q$ is positive definite. By further assuming that $Q$ is symmetric, i.e., $Q = Q^T$, are there any conditions on $A$ such that $A Q^{-1}A^T$ can be made either positive semi-definite or positive definite?

Rodrigo de Azevedo
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Guangyao
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  • Since any positive definite matrix $M \in \mathscr{M}n(\mathbb{R})$ (by definition symmetric) can be expressed as $M=PP^{\mathrm{t}}$ for a certain matrix $P \in \mathscr{M}_n(\mathbb{R})$, it follows that $RMR^{\mathrm{t}}=(RP)(RP)^{\mathrm{t}}$ will always be positive semi-definite for any $R \in \mathscr{M}{m,n}(\mathbb{R})$. If furthermore $m \leqslant n$ and $R$ is of rank $m$ one can show that $RMR^{\mathrm{t}}$ is in that case positive definite. – ΑΘΩ Mar 12 '21 at 11:51
  • Thanks for the prompt reply. 2 comments: A positive definite matrix may not be symmetric, see https://math.stackexchange.com/questions/1954167/do-positive-semidefinite-matrices-have-to-be-symmetric/1954174#1954174. Also, the second part on positive definiteness is not obvious. Can you give a more detailed explanation for that part? – Guangyao Mar 12 '21 at 13:17
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    @Guangyao The answer given there is a bit deceptive. There are many contexts in mathematics where positive semidefinite matrices are defined to necessarily be symmetric. For instance, wikipedia adopts this convention. – Ben Grossmann Mar 12 '21 at 17:34
  • @Guangyao I was considering the standard acceptation of positive definiteness which tacitly assumes symmetry. I agree that one could give a more general treatment of positive definiteness, without implicitly making this assumption (of symmetry). As to my second claim, returning to the notation used in my previous comment the claim amounts to stating that $(RP)(PR)^{\mathrm{t}}$ is positive definite when $m \leqslant n$ and $R$ is of rank $m$. Since $P \in \mathrm{GL}_n(\mathbb{R})$ is invertible (being that the originally given $M$ is invertible), we therefore have (to be cont.) – ΑΘΩ Mar 13 '21 at 03:40
  • @Guangyao (cont.) have that $\mathrm{rk}(RP)=\mathrm{rk}R=m$. By virtue of the Binet-Cauchy expansion of the determinant of $(RP)(PR)^{\mathrm{t}}$ we gather this determinant is expressed as a sum of squares, at least one of which nonzero (the square of one of the nonzero minors of order $m$ of $RP$). Thus, the matrix $RMR^{\mathrm{t}}=(RP)(PR)^{\mathrm{t}}$ is clearly semi-positive definite (including symmetry) and also invertible, hence being positive definite. – ΑΘΩ Mar 13 '21 at 03:45
  • @ΑΘΩ For the explanation of your second claim, do you mean $(RP)(RP)^T$ instead of $(RP)(PR)^T$? – Guangyao Mar 13 '21 at 15:59
  • @Guangyao Yes, I apologise for the mistake. I specifically mean to refer to the matrix $RMR^{\mathrm{t}}=RPP^{\mathrm{t}}R^{\mathrm{t}}=(RP)(RP)^{\mathrm{t}}$. – ΑΘΩ Mar 14 '21 at 06:18

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If $Q$ is positive semidefinite, then $AQ^{-1}A^T$ will always be positive semidefinite.

If $Q$ is positive definite, then it can be written in the form $$ Q = H + K, $$ where $H = \frac 12(Q + Q^T)$ is symmetric and positive definite and $K = \frac 12(Q - Q^T)$ is skew-symmetric. Because $H$ is positive definite, there exists an invertible matrix $P$ such that $S = PP^T$, and we have $$ P^{-1}QP^{-T} = P^{-1}HP^{-T} + P^{-1}KP^{-T} = I + P^{-1}KP^{-T} $$ where $I$ denotes the identity matrix. Note in particular that $J:= P^{-1}KP^{-T}$ is a skew-symmetric matrix. Note that $Q^{-1}$ is positive definite if and only if the matrix $$ P^TQ^{-1}P = [P^{-1}QP^{-T}]^{-1} = [I + J]^{-1} $$ is positive definite. Thus, we have reduced the question to that of whether/when $[I + J]^{-1}$ is positive definite for a skew-symmetric $J$.

Note that a matrix $M$ is positive definite if and only if the symmetric matrix $M + M^T$ is positive semidefinite. Taking $M = [I + J]^{-1}$, we have $$ M + M^T = [I + J]^{-1} + [I + J]^{-T} = [I + J]^{-1} + [I - J]^{-1}. $$ I claim that this can be rewritten as $$ [I + J]^{-1} + [I - J]^{-1} = 2[(I + J)(I - J)]^{-1} = 2 (I - J^2)^{-1}. $$ Now, we note that $I - J^2$ is necessarily positive definite because it is the sum of a positive definite matrix $I$ and a positive semidefinite matrix $-J^2 = JJ^T$. Thus, $(I - J^2)^{-1}$ and hence $(I + J)^{-1}$ is positive definite.

Ben Grossmann
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  • You make a series of important structural observations in your answer. Permit me a few comments: first of all, I believe the opening claim of your second paragraph means to say that $M$ is positive (semi)definite if and only if the symmetric part of $M$ is likewise positive (semi)definite. Second, adopting for arbitrary $M \in \mathscr{M}n(\mathbb{C})$ the definition that $M$ is positive definite if given any nonzero column matrix $X \in \mathscr{M}{n,1}(\mathbb{C})$, the unique element in the matrix $X^MX \in \mathscr{M}_{1}(\mathbb{C})$ (to be cont.)* – ΑΘΩ Mar 13 '21 at 04:33
  • (cont.) is a strictly positive real number, then it appears there is a considerably more direct way to establish the implication $M \in \mathrm{GL}_n(\mathbb{C})$ positive definite entails that $M^{-1}$ is likewise positive definite. Specifically, one considers arbitrary $X \in \mathrm{M}{n,1}(\mathbb{C})\setminus \left{\mathrm{O}{n,1}\right}$ and realises that $X^M^{-1}X=X^M^{-}M^M^{-1}X=Y^M^Y$, where $Y=M^{-1}X \neq \mathrm{O}_{n,1}$. The conclusion follows from the positive definiteness of $M^*$, which itself is an immediate consequence of the positive definiteness of $M$. – ΑΘΩ Mar 13 '21 at 04:38
  • @Ben Grossmann, I don't see a direct connection from the above analysis to the $A$ matrix given in the question. Is it possible to provide a more direct explanation/condition? – Guangyao Mar 13 '21 at 16:06