I'm working through the ideas in this document from Berkley. They claim (I'm paraphrasing), "$TT^* = T^*T$ if and only if $T$ is of the form $T = R + iM$ where $R$ and $M$ are self-adjoint." I'm sure you can prove this result by invoking the spectral theorem; but this document is about building intuition about how normal operators may have arisen, and how this intuition leads to the spectral theorem. So, I'm trying to prove this claim WITHOUT invoking the spectral theorem.
I've showed the reverse direction. In attempting to show the forward direction, I've first showed that any operator can be written $T = A + B$ where $A$ is self-adjoint and $B$ is skew self-adjoint, i.e. $B^* = -B$. Now, I want to show that $B$ is actually of the form $B = iC$ where $C$ is self-adjoint. Here is my attempt:
\begin{align*}
0 &= TT^* – T^*T \\
&= (A + B)(A + B)^* – (A + B)^*(A + B) \\
&= (A + B)(A^* + B^*) – (A^* + B^*)(A + B) \\
&= (AA^* – A^*A) + (AB^* – A^*B) + (BA^* – B^*A) + (B^*B – BB^*) \\
&= 0 -2AB + 2BA + 0.
\end{align*}
So, my attempt ends up proving that if $T$ is normal, then the self-adjoint and skew self-adjoint parts commute, which is neat. But it's not what I wanted!
Any ideas about what to try next? Thanks!
Best Answer
For any operator $T$ we can define $A=\frac {T+T^{*}} 2$ and $B=\frac {T-T^{*}} {2i}$. These are self-adjoint and $T=A+iB$. Not every operator $T$ is normal. So the statement you are trying to prove is false.