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!
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 $$A={1\over 2}(T+|T|),\quad B={1\over 2}(|T|-T)$$