In algebra, it has come up a number of times that if $A$ is a linear operator, then, for any integer $k>0$, $A^k$ inherits the eigenvectors of $A$. This is a very straightforward proof. However, $A$ does not necessarily share all eigenvectors of $A^k$. One obvious example of this is when $A^2=I$.
Another example I just encountered was in quantum mechanics, when the momentum operator $\hat{p}$ does not share all the eigenfunctions of $\frac{\hat{p}^2}{2m}$, the kinetic energy operator.
I can't seem to think of a finite-dimensional operator off the top of my head where $A$ does not share all the eigenvectors of $A^k$ and $A^k\not=I$, and was wondering if this has something to do with quantum mechanics being over an infinite dimensional vector space, something I have not encountered in my mathematics courses thus far.
Is it true that $A$ and $A^k\not=I$ always share eigenvectors when they are linear operators over a finite dimensional vector space? If so, what changes when the dimension becomes infinite?
I'm specifically asking about complex vector spaces, but would welcome answers about vector spaces over other fields as well.
For example, try $$ A = \pmatrix{1 & 0 & 0\cr 0 & -1 & 0\cr 0 & 0 & 0\cr}$$