Suppose S1 and S2 are self-adjoint operators in L(V ) such that: T = S1 + iS2. Show that S1 = T1 and S2 = T2 where T1 and T2

Question

We hare given that:

$T_1 := \frac{1}{2} (T + T^*)$ and $T_2 := \frac{1}{2i}(T + T^*)$

What I am mostly unsure of is my proof. Does it actually make any mathematical sense or if I have convinced myself of something false? It goes as follows:

Suppose that $S_1 \neq T_1$ and $S_2 \neq T_2$. Since $T=S_1 +iS_2$

\begin{split} T & \neq \frac{1}{2}(T + T^*) + \frac{i}{2i}(T + T^*) \\ & \neq \frac{1}{2}(T+T^*+T-T^*) \\ & \neq \frac{1}{2}(2T) \\ & \therefore T\neq T \end{split}

This is clearly not true, so the assumption that $S_1 \neq T_1$ and $S_2 \neq T_2$ leads to a contradiction that $T \neq T$. $\therefore S_1 = T_1$ and $S_2 = T_2$. $_\blacksquare$

Answer 1

No, this doesn't work. You can't add together non-equalities like this: $S_1\neq T_1$ and $S_2\neq T_2$ does not imply $S_1+iS_2\neq T_1+iT_2$, which is essentially what you are doing. For a simple example with numbers, consider that $1\neq 2$ and $4\neq 3$ but $1+4=2+3$.

Keep in mind also that your goal is to prove $S_1=T_1$ and $S_2=T_2$. The negation of this is $S_1\neq T_1$ or $S_2\neq T_2$, not $S_1\neq T_1$ and $S_2\neq T_2$ as you assumed.

Answer 2

$T^{*}=(S_1+iS_2)^{*}=S_1-iS_2$ because $(A+B)^{*}=A^{*}+B^{*}$ and $(cA)^{*}=\bar c A^{*}$. Now just use this equation and the equation $T=S_1+iS_2$ to calculate $S_1$ and $S_2$ in terms of $T$ and $T^{*}$.