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.

Suppose for a moment that we have just the real vector space $V = \mathbb{R}^2$. Here is an example to see why a geometric rotation will correspond to a complex eigenvalue.
First, for a concrete example, say $M$ is a rotation by $90$ degrees. If we only consider real numbers, we can easily check that $M$ has no real eigenvalues. But suppose we want to "pretend" $M$ had some eigenvalue, $\lambda$. We don't know yet that $\lambda$ is complex, just that it can't be real. 
We know there is some (nonzero) vector $x$ with $Mx = \lambda x$. Then $M^2x = \lambda^2 x$. But $M^2$ is a rotation by $180$ degrees, so $M^2 x = -x$. Hence $\lambda^2 = -1$. So $\lambda$ is a square root of $-1$ -- this immediately suggests we should look at complex numbers for $\lambda$, and in particular $\lambda$ should be be $i$ or $-i$.  
Similarly, say $N$ is a rotation by $60$ degrees, which also has no real eigenvalues. Then we will have $N^3 = -I$, and so any eigenvalue of $N$ should be a non-real cube root of $-1$. 
This is why rotations correspond with complex eigenvalues, in general: because iterated rotations go "around a circle" in the same way as iterated powers of complex numbers on the unit circle. 
