TITLE: Eigenvalues of the rank one matrix $uv^T$
QUESTION [59 upvotes]: Suppose $A=uv^T$ where $u$ and $v$ are non-zero column vectors in ${\mathbb R}^n$, $n\geq 3$. $\lambda=0$ is an eigenvalue of $A$ since $A$ is not of full rank. $\lambda=v^Tu$ is also an eigenvalue of $A$ since
$$Au = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ 
Here is my question:

Are there any other eigenvalues of $A$?

Added:
Thanks to Didier's comment and anon's answer, $A$ can not have other eigenvalues than $0$ and $v^Tu$. I would like to update the question:

Can $A$ be diagonalizable?

REPLY [11 votes]: The matrix $A=uv^T$ has rank$~1$, unless either $u$ or $v$ is zero, in which case $A=0$; assume the latter is not the case. By rank-nullity, $\ker(A)$ (the eigenspace of$~A$ for the eigenvalue$~0$) has dimension $n-1$, so $\lambda=0$ is a root of the characteristic polynomial $\chi_A$ with multiplicity at least$~n-1$. The sum of all such roots (counted with multiplicity) is $\def\tr{\operatorname{tr}}\tr A=v^T u$, so $\chi_A=x^{n-1}(x-\tr A)$. If $\tr A\neq0$ then the (direct) sum of the eigenspaces has dimension $(n-1)+1=n$, which implies that $A$ is diagonalisable. If however $\tr A=0$ then $0$ is the unique eigenvalue; but its eigenspace being of dimension only $n-1$, this means that $A$ is not diagonalisable in this case.
Another way to see that there are no eigenvalues other than $0$ and $c=v^T u$ is by doing the computation $(A-cI)\circ A=0=uv^Tuv^T-(v^Tu)uv^T$ since $v^T u$ is scalar. So eigenvalues of $A$ are roots of the polynomial $(X-c)X$; indeed the is the minimal polynomial of$~A$ (see this related answer).