Ok. I now feel this question is a little dumb. I finally know why. I hope my answer will be helpful for someone else.
The reason is so simple! Since we are trying to maximize ${{\bf{w}}^{\rm{T}}}{\bf{Sw}}$ and now we know the candidates are eigenvectors, just plug them back in, then we have ${{\bf{w}}^{\rm{T}}}{\bf{Sw}} = {{\bf{w}}^{\rm{T}}}\lambda {\bf{w}} = \lambda {{\bf{w}}^{\rm{T}}}{\bf{w}} = \lambda $, that means the eigenvalues are nothing but the variances. That's why we rank by eigenvalues!
It is so simple, but it really took me an entire day to figure out...
The $2 \times 2$ case is quite simple: such an $A$ causes a shear transformation: one vector is fixed, the one perpendicular to it maps to one with the same perpendicular component, but the parallel component is changed (think of the effect on the basis consisting of $(1,0)$ and $(0,1)$: $(1,0) \mapsto (1,0)$ and so is unchanged, whereas $(0,1) \mapsto (1,1)$.
The general case of only having one eigenvector is similar: in the two-dimensional case, the possible Jordan Normal Forms are
$$ \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}, \quad \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}, \quad \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, $$
where $\lambda \neq \mu$, and it is a significant theorem that any $2 \times 2$ matrix is related to one of these by a change-of-basis.
Now, the characteristic polynomial of any matrix of the first type is $(t-\lambda)(t-\mu)$, and its coefficients are real if the original matrix is real, so they are either both real or a complex conjugate pair. The complex case can be shown to be a rotation matrix, which has no real eigenvectors in two dimensions. The real case is a pair of unequal scalings.
The second case is a simple scaling, and it is easy to show that the only matrices of this type are of the form $\lambda I$ for real $\lambda$.
The last case is what your $A$ falls into, and it is easy to see that these are all shears, in the same way as $A$ is, but with additional scaling (since it factors into $\lambda I \begin{pmatrix} 1 & 1/\lambda \\ 0 & 1 \end{pmatrix}$).
Best Answer
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.