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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 $AB=0$.

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

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  • $\begingroup$ $T$ is self-adjoint and so, $T=T^*$ which means $|T|=T$. Did I lost somthing? $\endgroup$ – hamid kamali Apr 8 '16 at 20:14
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    $\begingroup$ @hamidkamali : In that case, $|T|=\sqrt{T^2}$ is the positive square root of $T^2$. $\endgroup$ – DisintegratingByParts Apr 8 '16 at 21:22
  • $\begingroup$ @TrialAndError How can we rigorously show $A$ and $B$ are positive? $\endgroup$ – Max Nov 11 '16 at 18:04
  • $\begingroup$ @Max : You are asking how to show that $-|T| \le T \le |T|$? $\endgroup$ – DisintegratingByParts Nov 11 '16 at 19:51
  • $\begingroup$ 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. $\endgroup$ – Max Nov 11 '16 at 19:54
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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?

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

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