linear algebra - What is the structure of a matrix when all the eigenvalues are complex?

Rotation matrices $\begin{bmatrix}\cos{\theta} & -\sin{\theta}\\\sin{\theta} & \cos{\theta}\end{bmatrix}$ or matrices with the structure $\begin{bmatrix}\sigma & \omega\\-\omega & \sigma\end{bmatrix}$ or $\begin{bmatrix}\sigma & -\omega\\\omega & \sigma\end{bmatrix}$ will always have complex eigenvalues.

Is the converse true i.e. will always the matrices with complex eigenvalues assume a structure like $\begin{bmatrix}\sigma & \mp{\omega}\\\pm{\omega} & \sigma\end{bmatrix}$? If so, then for any even order matrices such as $A$ with complex eigenvalues, should A always be a block diagonal with each diagonal matrix being in form as $\begin{bmatrix}\sigma & \mp{\omega}\\\pm{\omega} & \sigma\end{bmatrix}$?

Answer

I suppose that you assume that $A$ is real. If so then it has a real Schur form, that is, there exists a real orthogonal $Q$ and a block triangular $T$ such that

$$A=QTQ^T.$$
The matrices $T$ and $Q$ can be chosen such that $T$ has $1\times 1$ diagonal "blocks" corresponding to real eigenvalues and $2\times 2$ diagonal blocks of the form
$$\begin{bmatrix}\sigma&\omega\\-\omega&\sigma\end{bmatrix}, \quad \omega\neq 0,$$
corresponding to the conjugate pairs of complex eigenvalues $\sigma\pm i\omega$.

If, in addition, $A$ is normal ($AA^T=A^TA$), the matrix $T$ is block diagonal.

So, something like your claim is indeed true. It's just not that simple.