where V is a compelx inner product space prove that are selfadjoint?

Question

Let $\alpha \in End(V)$ where V is a complex inner product space. Define $$\alpha_1 = \frac{1}{2}(\alpha+\alpha^*)\\ and \\ \alpha_2 = \frac{1}{2i}(\alpha - \alpha^*)$$

(a) Prove that $\alpha_1$ and $\alpha_2$ are selfadjoint and that $\alpha = \alpha_1+i\alpha_2$.

(b) Suppose also that $\alpha = \beta_1 + i\beta_2$ where $\beta_1$ and $\beta_2$ are selfadjoint. Prove that $\beta_1 = \alpha_1$ and $\beta_2 = \alpha_2$

(c) Prove that $\alpha$ is normal if and only if $\alpha_1 \alpha_2 = \alpha_2 \alpha_1$

So from the definition of selfadjoint, $\alpha \in End(V)$ is selfadjoint if and only if $\alpha$ = $\alpha^*$. How would I start by showing that $\alpha_1$ and $\alpha_2$ are selfadjoint and then showing that $\alpha = \alpha_1+i\alpha_2$.

Okay with the help from the comments I was able to do parts (a) and (b).

Now I am only left to prove (c).

Answer 1

$$\alpha_1=\frac{1}{2}(\alpha+\alpha^*)$$ Then $$\alpha_1^*=\frac{1}{2}(\alpha^*+(\alpha^*)^*)=\frac{1}{2}(\alpha^*+\alpha)=\alpha_1$$ You can do the same for $\alpha_2$. For the rest, just multiply $\alpha_2$ by $i$, and then add $\alpha_1$

Answer 2

Some facts you need to prove first: $(cA)^{\ast}=c^{\ast}A^{\ast}$, $c^{\ast}$ is the complex conjugate of $c$, $(A^{\ast})^{\ast}=A$, $(A+B)^{\ast}=A^{\ast}+B^{\ast}$.