Normal linear maps over $\mathbb C$ 
Prove that if $\alpha: V \to V$ is a normal linear map on a finite-dimensional inner product space $V$ over $\mathbb C$ then $\alpha = \alpha_1 + i\alpha_2$ where $\alpha_1$ and $\alpha_2$ are self-adjoint, and $\alpha_1 \alpha_2 = \alpha_2 \alpha_1$.

Since every normal linear map is diagonalisable this is easy to show by diagonalising the matrix representing $\alpha$, but I have a feeling it can be shown directly from the definition of normal: $\alpha \alpha^* = \alpha^* \alpha$. Unfortunately I cannot see how to do it. A hint would be appreciated.

If we set
$\alpha_1 = \dfrac{\alpha + \alpha^\ast}{2}, \tag 1$
then evidently
$\alpha_1^\ast = \alpha_1, \tag 2$
i.e., $\alpha_1$ is self-adjoint; likewise, setting
$\alpha_2 = \dfrac{\alpha - \alpha^\ast}{2i}, \tag 3$
we see that
$\alpha_2^\ast = \dfrac{\alpha^\ast - \alpha}{-2i} = \dfrac{\alpha- \alpha^\ast}{2i}; \tag 4$
so $\alpha_2$ is also self-adjoint; also,
$\alpha = \alpha_1 + i \alpha_2, \tag 5$
as is easily seen.  Now
$\alpha_1 \alpha_2 = \dfrac{1}{4i}(\alpha + \alpha^\ast)(\alpha - \alpha^\ast) = \dfrac{1}{4i}(\alpha^2 - (\alpha^\ast)^2) = \dfrac{1}{4i}(\alpha - \alpha^\ast)(\alpha + \alpha^\ast) = \alpha_2 \alpha_1, \tag{6}$
where we have used $\alpha \alpha^\ast = \alpha^\ast \alpha$ in performing the algebraic operations of (6).  It follows that $\alpha_1$ and $\alpha_2$
meet the stated requirements.
No diagonalization needed!
