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

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.

• The way you've presented the question, $A$ and $B$ are obviously unique (for a given $T$) because they are defined by formulas. In order for the question of uniqueness to make sense, you need to ask why $R$ and $M$ are unique in the decomposition $T=A+iB$ of a normal operator $T$ into a combination of self-adjoint operators $R$ and $M$. And the answer as to why they're unique, is because you can solve for them in terms of $T$ (the aforementioned formulas).
– anon
Aug 12, 2022 at 3:28

$$A=\frac {T+T^{*}} 2$$ 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 $$T$$.

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.