# Compact operator space is the greatest ideal of $B(H)$

Suppose $H$ is a separable infinite dimensional Hilbert space. Show that if $A\in B(H)$ is noncompact, then there exist two operators $B,C$ such that $BAC=1$.

Clearly if $A$ is invertible it holds, but if $0\in \sigma(A)$, I do not have any idea. Please help me. Thanks.

Very rough outline:

• Suppose first that $A$ is self-adjoint.

• By the spectral theorem, we can write $A = \int \lambda dE$ where $E$ is the projection-valued spectral measure for $A$. Show that for sufficiently small $\epsilon$ we have that the projection $E([\epsilon, \infty))$ has infinite rank (if not, $A$ would be compact). Let $H_0$ be the image of this projection, which is a closed infinite-dimensional subspace of $H$. Note that $H_0$ is $A$-invariant, and that $\|Ax\| \ge \epsilon\|x\|$ for every $x \in H_0$.

• Since all infinite-dimensional separable Hilbert spaces are isometrically isomorphic, find an operator $C$ mapping $H$ isometrically onto $H_0$, and an operator $B_1$ mapping $H_0$ isometrically onto $H$.

• Now $B_1 A C$ is invertible. Taking $B = (B_1 A C)^{-1} B_1$ we are done.

• If $A$ is not self-adjoint, consider $A^* A$.

• It's a very nice proof. Thanks so much. – niki Jan 22 '15 at 13:49