# If $A\leq B$ in the sense of quadratic forms, then must $A^{-1} \geq B^{-1}$?

Let $A$ and $B$ be symmetric invertible operators on a Hilbert space $X$. Suppose $$\langle Ax , x \rangle \leq \langle Bx , x \rangle$$ for each $x\in X$. Does it follow that $\langle A^{-1} x ,x \rangle \geq \langle B^{-1} x , x \rangle$? What about in the finite dimensional case?

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For the finite-dimensional positive-definite case: math.stackexchange.com/questions/214126/… – Rahul Jan 13 '13 at 0:34
Doesn't it make a difference that the inner product is arbitrary? After all, $A^TM$ is not necessarily symmetric, and its inverse isn't $A^{-T}M$ – user7530 Jan 13 '13 at 6:42

It is not true even in the finite dimensional case. Let $A=A^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $B=B^{-1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. Then $\langle Ax,x\rangle\leq \langle Bx,x\rangle=\|x\|^2$ for all $x\in\mathbb C^2$, but with $x=(0,1)$ we get $\langle A^{-1}x,x\rangle=0\not\geq1=\langle B^{-1}x,x\rangle$.

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What if we assume the operators are positive definite? – user15464 Jan 13 '13 at 0:04

This is true if one assumes the operators to be positive, in the sense that $\langle Ax,x \rangle \geq 0$ for all $x \in X$. In fact, one has the following, more general, result:

Proposition. Let $\mathcal A$ be a $C^*$-algebra and $0 \leq a \leq b$ with $a,b\in \mathcal A$ and $a$ invertible. Then $b$ is invertible and $0 \leq b^{-1} \leq a^{-1}$.

Here, $a \geq 0$ means that $a$ is Hermitian and $\sigma(a) \subseteq [0,\infty)$, where $\sigma(a)$ is the spectrum of $a$. We write $a \geq b$ if $a-b \geq 0$. In the original problem, $\mathcal A$ would be the algebra of bounded operators on $X$ and positivity translates to the property $\langle Ax,x \rangle \geq 0$ for all $x \in X$.

The proof can be found here, Proposition 7.23 (it's in German, but I can translate it if there is any interest).

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