$\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?

  • $\begingroup$ I'd try one of those "other ways" myself. $\endgroup$ Sep 12, 2020 at 1:11
  • $\begingroup$ You can use the fact that the trace is cyclic and therefore invariant under basis change. So you might as well start with $N$ being diagonal. $\endgroup$
    – abhi01nat
    Sep 12, 2020 at 6:08
  • $\begingroup$ Your equation on the eigenvalues does not generally hold. We can guarantee that $A$ can be decomposed into $A = B + iC$ with this relationship between the eigenvalues whenever $A$ is normal, but not in the general case. $\endgroup$ Sep 12, 2020 at 17:44
  • $\begingroup$ @BenGrossmann Eq. (3) holds in general, right? If it does, then eq. (4) has to hold too because $\sum_k \beta_k + i \sum_l \gamma_l = \text{tr} \left( A \right) = \sum_m \alpha_m$ Unless I'm missing something? $\endgroup$
    – Attila Kun
    Sep 12, 2020 at 20:48
  • 1
    $\begingroup$ @AttilaKun Sorry, I had read that incorrectly: I thought you meant $\alpha_k = \beta_k + i \gamma_k$ $\endgroup$ Sep 12, 2020 at 20:52

1 Answer 1


For $A\in\mathbb{C}^{n\times n}$, its characteristic polynomial can be writen as $$|\lambda I-A|= \begin{vmatrix} \lambda-x_{11} & x_{12} & \cdots & x_{1n}\\ x_{21} & \lambda-x_{22} & \cdots & x_{2n}\\ \vdots & \vdots &        &\vdots\\ x_{n1} & x_{n2} & \cdots & \lambda-x_{nn}\\ \end{vmatrix}=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdots(\lambda-\lambda_l) $$

compare the coefficient of $\lambda^{n-1}$.

  • 1
    $\begingroup$ This does not answer the question being asked $\endgroup$ Sep 12, 2020 at 17:40

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