The geometric interpretation of real eigenvalues and their eigenvectors seems straightforward – apply A to one of its eigenvectors, and you get back a parallel vector. However, I did not (and perhaps still don’t) understand from a geometric perspective why the eigenvectors of $A$ and $A^2$ must be the same. I understand why from an algebraic perspective this is the case, but geometrically why couldn’t there be a vector such that when you apply $A$ to it once, it goes off in some other direction; but upon applying $A$ to it a 2nd time, it is back in its original direction?
It then dawned on me that this must be what imaginary eigenvalues are telling us – an eigenvalue with imaginary component $i$, for example, tells us that the associated eigenvector will return a vector with parallel real components (times some constant) for all even powers of $A$. And if the imaginary component were, say, $i^{1/3}$ it would tell us that every $A$ to a multiple of 3 returned a vector with a parallel real component. That is, in the complex world, eigenvectors are always parallel. But geometrically, the imaginary component of the eigenvalue is telling us that the real component of the eigenvector is parallel to the original real component of the eigenvector with some frequency corresponding to the imaginary component of the eigenvalue. Is this correct?
Even if this is correct, I’m still confused. Let’s say we have an eigenvalue of $i^{1/3}$. It seems like there should be 3 vectors that all have this eigenvalue due to the fact that the real component of $A$’s eigenvectors are cyclic with period 3, so to speak. And since its complex conjugate will also be an eigenvalue, wouldn’t we actually have 6 eigenvectors with complex eigenvalues in this example – 3 rotated $i^{1/3}$ and 3 more rotated $i^{-1/3}$? None of this seems right, so I'm clearly not thinking about this correctly.
Thanks for any insight you can provide.
Best Answer
An Eigenvector is such that $$Av=\lambda v$$ where $\lambda$ is a scalar.
Unavoidably, if you iterate $A$,
$$A(Av)=A(\lambda v)=\lambda^2v=\lambda'v.$$
This works for real and complex $\lambda$. But even with $\lambda$ complex, $\lambda v$ is parallel to $v$.