Eigenvalues of a traceless matrix

The trace of a $n \times n$ matrix $M$ is the sum of its eigenvalues, which itself is the opposite of the coefficient of $x^{n-1}$ in its characteristic polynomial $\chi_M(\lambda)$.

Therefore, using Cayley-Hamilton theorem (saying that $\chi_M(M)=0)$, one can characterize a traceless matrix as a matrix whose $n$th power can be expressed as a linear combination of $I,M,M^2,...M^{n-2}$ (no $M^{n-1}$).

In the particular case $n=3$, the condition boils down to the existence of constants $p,q$ such that :

$$\chi_M(M)=0 \ \ \iff \ \ M^3-pM-qI_3=0 \ \ \iff \ \ M^3=pM+qI_3\tag{1}$$

with necessarily $q=\det(M)$ (once again due to Cayley-Hamilton theorem).

An illustrative example : let us consider the case where $M$ is a $3 \times 3$ skew-symmetric (hence traceless) matrix :

$$M=\begin{pmatrix}\ \ \ 0&-c& \ \ \ b\\ \ \ \ c& \ \ \ 0&-a\\-b& \ \ \ a& \ \ \ 0\end{pmatrix}$$

A small computation shows that

$$M^3=-(a^2+b^2+c^2)M+0 I_3\tag{2}$$

(compare with formula (1))

Geometric interpretation : such a matrix can be associated with the linear operation "cross product with vector $V=(a,b,c)$" often denoted $[V]_{\times}$ (see here). Therefore, formula (3) means that, for any vector $W$

$$V \times (V \times (V \times W))= -\|V\|^2 W$$

(make a drawing !)

Suppose we're working in $\mathbb{C}$, such that there're always n eigenvalues counting multiplicity.

In 3D, you can simply require the matrix to be not invertible (so $0$ is an eigenvalue). Then by $\lambda_1+\lambda_2+\lambda_3=0$, the eigenvalues look like $(0,\lambda,-\lambda)$. However, for higher dimensions, simple conditions like the 3D case feels unlikely to exist for me.

You can add strong(special) conditions on $M$ to make it true though. For example: any matrix that is similar to some real skew-symmetric matrix is traceless, and it has purely imaginary eigenvalues that do come in pairs.