$T$ 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 $AB=0$. 

 A: 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?
A: 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
$$
             A = \frac{1}{2}(|T|+T),\;\;\; B=\frac{1}{2}(|T|-T), \\
                     |T| = (T^{\star}T)^{1/2}
$$
