Why do complex eigenvalues correspond to a rotation of the vector? We have a linear transformation $T: \Bbb R^m \to \Bbb R^n$ defined by $T(x)=Ax$ for $x \in \Bbb R^m$ and $A \in M_{n \times m}(\Bbb R)$.  I understand why real-valued eigenvalues of $A$ correspond to scaling the length of the associated eigenvectors, but why is it that complex eigenvalues are said to rotate the eigenvector?
If you have an eigenvector $x = (x_1, x_2, x_3)$ whose eigenvalue is $\lambda = a+bi$, how is $\lambda x = ((a+bi)x_1, (a+bi)x_2, (a+bi)x_3)$ a rotation of $x$?  Shouldn't a rotation just be something like $\sin$'s and $\cos$'s multiplied by each component?  Where do the imaginary parts fit in?
Thanks.

For a real matrix, if $a + bi$ is an eigenvalue, so is $a - bi$. And in fact, the classic matrix with this pair of eigenvalues is 
$$
\begin{bmatrix}
a & -b \\
b & a
\end{bmatrix}.
$$
You probably know some theorems that say you can sometimes diagonalize a matrix by changing basis. In the cases where diagonalization isn't possible (for real matrices), which you can get is $2 \times 2$ blocks like that one on the diagonal. 
So what's that $2 \times 2$ matrix do? Well, it multiplies lengths by $r = \sqrt{a^2 + b^2}$. So let's assume we've already factored that out into a diagonal matrix:
$$
\begin{bmatrix}
a & -b \\
b & a
\end{bmatrix} = \begin{bmatrix}
r & 0 \\
0 &r
\end{bmatrix} \cdot 
\begin{bmatrix}
a' & -b' \\
b' & a'
\end{bmatrix}
$$
so that $a'^2 + b'^2 = 1$. Letting $\theta = \arctan(b'/a')$, we get $\sin \theta = b', \cos \theta = a'$, so this remaining matrix is just a rotation by angle $\theta$. 
Does that help? 
