$A$ and $B$ are p.d,show that $A$ - $B$ is p.s.d iff $B^{-1}$ - $A^{-1}$ is p.s.d

Question

$A$ and $B$ are p.d,show that $A$ - $B$ is p.s.d iff $B^{-1}$ - $A^{-1}$ is p.s.d

I have no idea how to prove it.Any hint? Thanks in advance!

Answer 1

Since $A$ has full rank you may as well replace $A$ and $B$ with $I$ and $A^{-1/2}BA^{-1/2}$ by a change of coordinates. Let $C=A^{-1/2}BA^{-1/2}$, let its eigenvalues be $c_i$. Now the assumption is that the eigenvalues of $I-C$ are all nonnegative, and you want to conclude that those of $C^{-1}-I$ are also all nonnegative. That is, from $1-c_i\ge0$ you want to deduce $c_i^{-1}-1\ge0$.

Answer 2

For fun, here is the sketch of a deliberately long-winding answer. For every vector $x$, let $$ y=A^{-1/2}(A^{-1/2}BA^{-1/2})^{-1/2}A^{-1/2}x. $$ One may verify that $$ y^\ast Ay=x^\ast B^{-1}x,\quad y^\ast By=x^\ast A^{-1}x. $$ Therefore $y^\ast(A-B)\,y\ge0$ if and only if $x^\ast(B^{-1}-A^{-1})\,x\ge0$. Since $x\mapsto y$ is bijective, we conclude that $A-B$ is PSD if and only if $B^{-1}-A^{-1}$ is PSD.

Answer 3

First we will show that if $(A-B)$ is psd, then $(B^{-1} - A^{-1})$ is also psd. Then we will do it the other way round.

Now let $A - B$ be psd. Then $x^\top (A-B) x \ge 0$, which implies that $x^\top Ax \ge x^\top Bx$ Note that $x^\top A x$ is a scalar and so is the other term and they are both positive because A and B are pd. Now let $x^\top Ax$ be a scalar $k$, and let $x^\top Bx$ be p. $$k \ge p$$ $$1/k \le 1/p$$

Now since, $k = x^\top Ax$ it follows that $1/k = x^\top A^{-1}x$ $$x^\top A^{-1} x \le x^\top B^{-1} x$$ $$x^\top (B^{-1} - A^{-1}) x \ge 0$$ which shows that $(B^{-1} - A^{-1})$ is psd.

Now we can follow the same pattern and prove it the other way. Don't forget that for an iff proof you need to work in both directions!