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 $\begin{bmatrix}\sigma & \mp{\omega}\\\pm{\omega} & \sigma\end{bmatrix}$? 
 A: A real $2\times 2$ matrix has complex (non-real) eigenvalues if and only if $a_{21}a_{12}<-\left(\frac{a_{11}-a_{22}}2\right)^2$. This inequality comes out by evaluating the sign of the discriminant of the characteristic polynomial $p(t)=\det(A-tI)$. So the claim does not hold. For one thing, you may change $a_{12}$ and $a_{21}$ accordingly while preserving their product. 
A: 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.
