decomposition of an operator on a finite-dimensional complex inner product space

Question

For any operator $T$ we can define $A=\frac {T+T^{*}} 2$ and $B=\frac {T-T^{*}} {2i}$.

So $T$ can be written as $T = A + iB$ and $A,B$ are two self-adjoint operators.

Now, suppose an operator $T$ is written in the form $T = R + iM$, where $R, M$ are two self-adjoint operators. Do $R$ and $M$ have to be $A$ and $B$? In other words, is the decomposition unique?

This comes from an answer to this question.

Answer 1

$A=\frac {T+T^{*}} 2$ is self adjoint operator.

$iB=\frac {T-T^{*}} 2$ skew adjoint operator.

Claim : $\mathcal{L}({\mathcal{H}})=\mathcal{S}(\mathcal{H})\oplus \mathcal{SK}({\mathcal{H}})$

1. $T=A+iB$

2. $T\in \mathcal{S}(\mathcal{H})\ \cap\mathcal{SK}({\mathcal{H}})$ implies $T^{\star}=T$ and $T^{\star}=-T$.Hence $2T=0$ implies $T=0$.

Hence the pair $(A, B) $ is unique for $T$.

Answer 2

I believe it should be unique. Correct me if I am wrong, but you can just "construct" $A$ and $B$ as you want them to be. That is suppose we have that $T=A+Bi$ is this decomposition of $T$ into a sum of two self-adjoint operators, then we have that $T^*=A-Bi$. Thus, if we calculate $\frac{T+T^*}{2}=\frac{A+Bi+A-Bi}{2}=A$. Similarly, we have that $\frac{T-T^*}{2}=B$. Thus, we have that this decomposition should be unique.