Relation between the eigenvalues of a matrix A and the eigenvalues of its hermitian and skew-hermitian parts It is well know that every matrix $A\in M_n(\mathbb{C})$ can be decomposed in a unique way as a sum of its hermitian part $H$ and its skew-hermitian part $K$:  $$A=\frac{A+A^*}{2}+\frac{A-A^*}{2}$$
However, there is a relationship between the eigenvalues of $A$, $H$ and $K$ that I´m trying to prove without success:
If the spectrums of $A, H, K$ are $\sigma(A)= \{ \lambda_1,\ldots,\lambda_n  \}$, $\sigma(H)= \{ \alpha_1,\ldots,\alpha_n \}$, $\sigma(K)= \{ \beta_1, \ldots, \beta_n  \},$ then $$\sum_{i=1}^n\mathopen|\lambda_i\mathclose|^2\leqslant \sum_{i=1}^n\mathopen|\alpha_i\mathclose|^2+ \sum_{i=1}^n\mathopen|\beta_i\mathclose|^2$$ With equality $\text{iff}$ $A$ is normal.

Note: My convention for $H$ and $K$ is to take $H = \frac{A + A^*}{2}$ and $K = \frac{A - A^*}{2i}$, so that $A = H + iK$.  With this convention, $H$ and $K$ are both Hermitian.  Notably, they have real eigenvalues.
First, note that
$$
\DeclareMathOperator{\tr}{tr}
\sum |\lambda_i|^2 \leq \tr(A^*A)
$$
Something along these lines is usually proven together with the spectral theorem for normal matrices following Schur's theorem (see for example Horn and Johnson). Notably, the inequality becomes equality if and only if $A$ is normal.
Now, expand $\tr(A^*A)$ in terms of $H$ and $K$ to get
$$
\tr(A^*A) = 
\tr[(H+iK)^*(H+iK)]=
\tr(H^*H)+ i\tr(HK) - i \tr(HK)+\tr(K^*K)=\\
\tr(H^*H)+ \tr(K^*K)
$$
Note, however, that both $H$ and $K$ are normal.
Or, a yet more straightforward observation at this point is that since both $H$ and $K$ are Hermitian, we have
$$
\tr(H^*H)+ \tr(K^*K) = 
\tr(H^2)+ \tr(K^2) = \sum_{i} \alpha_i^2 + \sum_i \beta_i^2
$$
where we note that these eigenvalues must be real.
