I would like to prove the following claim:
Suppose A and B are (square, symmetric) positive definite matrices of dimension nxn (with underlying vector space V).
Then, B-A is positive semi-definite IFF $A^{-1}$ – $B^{-1}$ is positive semi-definite.
My attempted approach is to use the variational characterisation of eigenvalues:
$\forall x \in V , x^T(B-A)x \geq 0 \Rightarrow x^TBx \geq x^TAx$
By the variational characterisation of eigenvalues of symmetric matrices,
$\lambda_k(B) = max_{dim(U) = k} min_{x\in U, ||x|| = 1} \ x^TBx \geq max_{dim(U) = k} min_{x\in U, ||x|| = 1} \ x^TAx = \lambda_k(A)$
$\forall \ \ 1\leq k\leq n$
If this is true, I think the claim follows just by comparing magnitudes of the eigenvalues of the inverses because A and B are Positive definite, but I am wondering whether the last step above is in fact correct. Does the maxmin above preserve $x^TBx \geq x^TAx$?
Edit: I am not sure if this actually helps with proving the main claim. Any hints?
Best Answer
lemma:
for real symmetric $C\succ \mathbf 0$
$I-C\succeq \mathbf 0$ iff $C^{-1}-I\succeq \mathbf 0$
the LHS is true iff all $\lambda^{(C)}\in (0,1]$ and the RHS is true iff all $\lambda^{(C^{-1})}\in [1,\infty)$, but $\big(\lambda^{(C)}\big)^{-1} = \lambda^{(C^{-1})}$ so one condition implies the other
main argument
select $C:= B^\frac{-1}{2}AB^\frac{-1}{2}$
$B-A = B^\frac{1}{2}\big(I -B^\frac{-1}{2}AB^\frac{-1}{2}\big)B^\frac{1}{2} = B^\frac{1}{2}\big(I -C\big)B^\frac{1}{2}$
$A^{-1} -B^{-1} = B^\frac{-1}{2}\big(B^\frac{1}{2}A^{-1}B^\frac{1}{2} - I\big)B^\frac{-1}{2}= B^\frac{-1}{2}\big(C^{-1} - I\big)B^\frac{-1}{2}$
recall that congruence implies same signature and notice
$B^\frac{1}{2}\big(I -C\big)B^\frac{1}{2}$ is congruent to $\big(I-C\big)$ and
$B^\frac{-1}{2}\big(C^{-1} - I\big)B^\frac{-1}{2}$ is congruent to $\big(C^{-1} - I\big)$
and by application of the lemma, the former is PSD iff the latter is