$\newcommand{\bra}[1]{\left<#1\right|} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\braket}[2]{\left<#1 | #2\right>} \newcommand{\tr}[1]{\text{tr}\left(#1\right)} $There are other ways of doing this, but out of curiosity I'm trying to use the spectral decomposition theorem to prove the statement in the title.

Due to the spectral decomposition theorem, a normal matrix $N$ can be written as: $$ N = \sum_j \nu_j \ket{w_j}\bra{w_j} $$ where $\nu_j$s are the eigenvalues and $\ket{w_j}$s are their associated eigenvectors.

This can be used to prove that the trace of $N$ equals the sum of its eigenvalues: $$ \tr{N} = \sum_i \bra{i} N \ket{i} = \sum_i \bra{i} \left( \sum_j \nu_j \ket{w_j}\bra{w_j} \right) \ket{i} = \sum_j \nu_j \sum_i \braket{i}{w_j} \braket{w_j}{i} = \sum_j \nu_j \braket{w_j}{w_j} = \sum_j \nu_j \tag{1}\label{1} $$

We also know that any linear operator $A$ can be written as:

$$ A = B + iC \tag{2}\label{2} $$

where $B$ and $C$ are Hermitian matrices. Hermiticity implies normality, so the spectral decomposition and \eqref{1} can be used on them to write:

$$ \tr{A} = \tr{B + iC} = \tr{B} + i\tr{C} = \sum_k \beta_k + i \sum_l \gamma_l \tag{3}\label{3} $$

where $\beta_k$ and $\gamma_l$ are the eigenvalues of $B$ and $C$ respectively. This is quite promising but we still need to prove that

$$ \sum_k \beta_k + i \sum_l \gamma_l = \sum_m \alpha_m \tag{4}\label{4} $$

where $\alpha_m$ are the eigenvalues of $A$. Is there a way to do this?