# $T$ is self-adjoint $\Rightarrow \exists$ positive $A,B$ such that $T=A-B$ and $AB=0$

I have a trouble by the following problem and I dont have any idea to solve it. can anybody give me a hint? Thanx in advance.

Let $\mathcal H$ be a Hilbert space and $T:\mathcal H \to \mathcal H$ be a self-adjoint operator. Show that there exists positive operators $A$ and $B$ suchthat $T=A-B$ and $AB=0$.

For real $t$, you have $$t = \frac{1}{2}(|t|+t)-\frac{1}{2}(|t|-t), \\ |t|+t \ge 0,\;\;\; |t|-t \ge 0,$$ and $$(|t|+t)(|t|-t) = |t|^2-t^2=0.$$ By analogy, try $$A = \frac{1}{2}(|T|+T),\;\;\; B=\frac{1}{2}(|T|-T), \\ |T| = (T^{\star}T)^{1/2}$$

• $T$ is self-adjoint and so, $T=T^*$ which means $|T|=T$. Did I lost somthing? – hamid kamali Apr 8 '16 at 20:14
• @hamidkamali : In that case, $|T|=\sqrt{T^2}$ is the positive square root of $T^2$. – DisintegratingByParts Apr 8 '16 at 21:22
• @TrialAndError How can we rigorously show $A$ and $B$ are positive? – Max Nov 11 '16 at 18:04
• @Max : You are asking how to show that $-|T| \le T \le |T|$? – DisintegratingByParts Nov 11 '16 at 19:51
• Yes. I can do it, say, in the case $T$ is compact (with a dense span of eigenvalues) but I have not been able to do it in general. – Max Nov 11 '16 at 19:54

I'm going to assume $\mathcal H$ is separable (so we have a countable basis) and $T$ is compact. Since $T$ is compact and self-adjoint, there's an orthonormal basis of $\mathcal{H}$ consisting of eigenvectors of $T$. Let this basis be $(e_i)_{i \in \mathbb N}$, satisfying $T e_i = \lambda_i e_i$. Let $$A e_i = \begin{cases} \lambda_i e_i & \lambda_i > 0 \\ 0 & \lambda_i \leq 0 \end{cases}, \qquad B e_i = \begin{cases} 0 & \lambda_i > 0 \\ - \lambda_i e_i & \lambda_i \leq 0. \end{cases}$$ Essentially, $A =T$ on basis vectors with positive eigenvalue and $A = 0$ otherwise; similarly, $-B = T$ on basis vectors with negative eigenvalue and $B = 0$ otherwise.

Can you verify that $A$ and $B$ have the desired properties?

• Yes, sorry, i should have added a negative sign on $B$; fixing now! – Jon Warneke Apr 8 '16 at 15:39