# Representations of Self Adjoint operators

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.

Thanks!

• 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. Commented Oct 30, 2022 at 21:57
• What things have you learned about square roots and positive operators. Commented Oct 30, 2022 at 23:05
• @Mason Unfortunately, spectral theorem is not covered in my notes.
– user775699
Commented Oct 31, 2022 at 0:41

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 $$A={1\over 2}(T+|T|),\quad B={1\over 2}(|T|-T)$$
• Hi! Can you please tell how using the 1st two lines you wrote $BA= B' A'=0$?
• $0=(AB)^*=B^*A^*=BA.$ Commented Dec 13, 2022 at 0:07