1

Given two $\textbf{positive definite (pd), Hermitian}$ matrices X and Y, I am trying to determine whether $(X-Y)(Y^{-1}-X^{-1})$ will always be pd as well, and how to prove this.

This formulation appears when trying to prove that the inverse function is convex on pd matrices: Is inverse matrix convex?.

It was trying to prove that if $(X-Y)\succ 0$, then $(Y^{-1}-X^{-1})\prec 0$ and vice versa, and subsequently $(X-Y)(Y^{-1}-X^{-1})\succ 0$ always. However I wasn't able to prove either of these two steps - nor disprove either of them.

Community
  • 1
vidit
  • 113
  • What "pd" does mean? –  May 18 '16 at 12:20
  • positive definite, I assume. – Michael Grant May 18 '16 at 13:47
  • This product isn't symmetric. Presumably $X$ and $Y$ are, since usually that is assumed when discussing positive definiteness. Can you provide more context, please? Where did this come from, and what have you tried so far? – Michael Grant May 18 '16 at 13:49
  • Updated the description to better define the problem – vidit May 18 '16 at 17:51
  • I'd still like you to define exactly what you mean with regards to the product---which is not Hermitian. Yes, I know positive definiteness is defined even for nonsymmetric matrices, but I think it will be helpful for you to establish your definition of positive definiteness in the non-symmetric case. – Michael Grant May 18 '16 at 23:48
  • 1
    This is readily refuted a number of ways. First of all, if X=Y, then the quantity is zero, as Luca points out below. But you can also try a variety of randomly generated matrices. – Michael Grant May 22 '16 at 20:50

1 Answers1

2

[Edited after After Michael's comment] In the general case, the OP's claim is not verified. For example if $$ X = \begin{bmatrix} 29& 11& 19\\ 11& 10& 12\\ 19& 12& 19 \end{bmatrix} , \quad Y = \begin{bmatrix} 14& 16& 17\\ 16& 21& 22\\ 17& 22& 29 \end{bmatrix} , \quad w = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} $$ we have that $$w^T(X-Y)(Y^{-1}-X^{-1})\,w = -21002/11025.$$

The claim is verified, instead, in the special case that $X$ and $Y$ are such that $XY^{-1}$ is symmetric but one needs to relax the claim from positive definite to positive semi-definite. In fact, if $X=Y=I$, the results is the matrix of all zeros.

We could rewrite $(X-Y)(Y^{-1}-X^{-1})$ as $XY^{-1}+YX^{-1}-2I$. Then, we observe that $XY^{-1} = (YX^{-1})^{-1}$, which implies that the two must have the same eigenvectors with reciprocal eigenvalues. Therefore, the eignevalues of $XY^{-1}+YX^{-1}$ will be $\lambda_i + 1/\lambda_i$, where $\lambda_i$ is the $i$-th eigenvalue of $XY^{-1}$. We observe that $x + 1/x \ge 2\; \forall x>0$, which implies that all eigenvalues of $XY^{-1}+YX^{-1}$ are at least 2. If $XY^{-1}$ is symmetric, this implies that $XY^{-1}+YX^{-1}-2I$ is at least positive semi-definite.

Luca Citi
  • 1,879
  • 14
  • 14
  • Unfortunately, $(X-Y)(Y^{-1}-X^{-1})$ is not necessarily symmetric. The relationship between nonnegative eigenvalues and positive semidefiniteness applies only to symmetric matrices. A nonsymmetric matrix $A$ is positive semidefinite if and only if the eigenvalues of $A+A^T$ are nonnegative. – Michael Grant May 22 '16 at 20:44
  • Thanks Michael for pointing this out. I have amended the answer and provided a counter example for the case when $XY^{-1}$ is not symmetric. – Luca Citi May 23 '16 at 09:58
  • Are you sure X and Y are psd? – Michael Grant May 23 '16 at 11:15
  • @MichaelGrant, you mean $X$ and $Y$ in my counter example? They were built as products of matrices with their transposes, e.g. $X=A^T A$ with $A=[[4, 0, 1],[2, 1, 3],[3, 3, 3]]$. – Luca Citi May 23 '16 at 12:50
  • Sounds good. They didn't look PD but my intuition isn't perfect! – Michael Grant May 23 '16 at 13:08