Let $u,v \in \mathbb{R}^n$ and let $A = u v^T$.
What is the largest eigenvalue of $A$ and what is the corresponding eigenvector?
It is easy to write down the definition of an eigenvalue/vector, but I'm not sure what to do from there.
I think an interesting observation is that
$$Au = u(v^Tu) = \lambda u$$
So perhaps $u$ is an eigenvector with eigenvalue $v^Tu$ (or perhaps that isn't the way outer products work).
Any hints are appreciated.
Best Answer
Some other eigenvalues/eigenvectors: what happens when you compute $Aw$ where $w$ is orthogonal to $v$?
Your observation is also correct. If $v^\top u \ne 0$, then this gives you a nonzero eigenvalue, and (by combining with the previous paragraph) you have then characterized all eigenvalues of $A$. If $v^\top u = 0$ however, then the only eigenvalues are $0$.