The statement that $A \ge B$ implies $A^{-1} \le B^{-1}$ is still true for matrices? [closed]

Question

Problem:

Suppose we have two real, symmetric and positive definite square matrices $A$ and $B$, i.e., $$A_{ij}, B_{ij}\in \mathbb{R}$$ $$A^T=A$$ $$B^T=B$$ $$x^TAx>0 \forall x$$ $$x^TBx>0 \forall x$$

If $A \ge B$, i.e., $A-B$ is semi-postive definite ($x^T(A-B)x \ge0,\,\forall x $), then is the statement that $A^{-1} \le B^{-1}$ true, i.e. $A^{-1}-B^{-1}\ge0$ ($x^T(B^{-1}-A^{-1})x\ge0,\,\forall x $)

Remarks:

Obviously if $A$ and $B$ can be diagonalized simultaneously with the same similarity transformation, then the statement is true. What about the general case? I tried some numerical examples, it seems the statement is true. But I don't know how prove it.

I would appreciate if anyone can give a proof or point out any reference that has the solution of the above problem.

Thanks in advance!

Answer 1

This is a well-known fact. A simple proof : setting $y=B^{1/2}x$, we have $\|y\|^2\le y^TB^{-1/2}AB^{-1/2}y$, that is $I_n\le B^{-1/2}AB^{-1/2}$. The eigenvalues of the latter symmetric matrix are thus $\ge1$. Its inverse $B^{1/2}A^{-1}B^{1/2}$ has eigenvalues $\le1$, that is $B^{1/2}A^{-1}B^{1/2}\le I_n$. This gives $z^TB^{1/2}A^{-1}B^{1/2}z\le\|z\|^2$. Setting $w=B^{1/2}z$, this writes $w^TA^{-1}z\le z^TB^{-1}z$, that is $A^{-1}\le B^{-1}$.

More generally, Loewner theory tells you what are the operator-monotone functions. By definition, a numerical function $f:I\rightarrow{\mathbb R}$ ($I$ an interval) is monotone operator if whenever $A,B$ are real symmetric matrices with spectra included in $I$, the inequality $A\le B$ implies $f(A)\le f(B)$. The beautiful theorem is that $f$ is operator monotone if and only if it admits a holomorphic extension to the upper half plane $\Im z>0$, with values in the upper half plane (of course this extension is unique) ; such a holomorphic function is called a Pick function. In particular, operator monotone functions are analytic ! For instance, if $\alpha>0$, the map $A\mapsto A^\alpha$ is monotone operator over $(0,+\infty)$ if and only if $\alpha\le1$ !!

W. F. Donoghue dedicated a full book to Loewner theorey. See also R. Bhatia's book.