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.

Basically because Complex multiplication is equivalent to scaling and rotation. I think writing Complex numbers in polar form is the easiest way of seeing this.
Assume $$ Av = \lambda v ; \lambda \in \mathbb C$$
Now, we can write both $$ v, \lambda $$ in polar form , as , respectively
$$re^{i \theta}, r'e^{ i\theta'} $$
Then $$\lambda v = rr'e^{i(\theta + \theta' )} $$
So when multiplying $$v:=re^{i\theta} $$ we end up with
$$rr'e^{i ( \theta + \theta')} $$ which is $$v$$ stretched by a factor$$r'$$
And rotated by an angle $$ \theta' $$
