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Consider this question asked in my mid term exam:

Let H be a Hilbert space. Show that if $T\in L(H)$ is self-adjoint, then T can be written uniquely in the form T=B-A , where A and B are +ve and $AB=BA=0.$

Attempt: I am not sure which result to use to prove that T can be written uniquely in the form of T=B-A.

To prove uniqueness , I let that T can be written as both B- A and B'- A' where A and B are positive in the hope of proving that B=B' and A=A' but I got struck on this as well.

So, can you please help me with this question?

Thanks!

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  • $\begingroup$ This is easy when $T$ is a diagonal matrix. This solves the case $H = \mathbb{R}^n$, due to the spectral theorem. In infinite dimensions there is also a spectral theorem. $\endgroup$
    – Mason
    Commented Oct 30, 2022 at 21:57
  • $\begingroup$ What things have you learned about square roots and positive operators. $\endgroup$ Commented Oct 30, 2022 at 23:05
  • $\begingroup$ @Mason Unfortunately, spectral theorem is not covered in my notes. $\endgroup$
    – user775699
    Commented Oct 31, 2022 at 0:41

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Concerning uniqueness no spectral theorem is needed. Assume $ AB=A'B'=0$ and $$T=A-B=A'-B'\quad (*)$$ Then $BA=B'A'=0$ and $$T^2=(A+B)^2=(A'+B')^2$$ In view of the uniqueness (see this) we obtain $A+B=A'+B'.$ Now $(*)$ implies $A=A'$ and $B=B'.$

The existence requires the knowledge of the square root of a nonnegative operator (no spectral theory needed). Then we define $|T|=(T^2)^{1/2}$ and $$A={1\over 2}(T+|T|),\quad B={1\over 2}(|T|-T)$$

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  • $\begingroup$ Hi! Can you please tell how using the 1st two lines you wrote $BA= B' A'=0$? $\endgroup$
    – user775699
    Commented Dec 12, 2022 at 21:36
  • $\begingroup$ $0=(AB)^*=B^*A^*=BA.$ $\endgroup$ Commented Dec 13, 2022 at 0:07

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