I've came across this issue, and couldn't manage to explain it.
If given a square matrix, such as $3\times 3$, and it's rank equals to $1$, when searching its eigenvalues, the Trace of the matrix is one of them, as well as $0$.
I would like to get the explanation of this issue and to ask if another eigenvalues could be possible (except the trace and $0$)?
EDIT: i mean that if a matrix is in its diagonal form it's pretty obvious, but if a matrix is not in its diagonal form or even it's unknown if the matrix is diagonalizable.
Best Answer
Say $u$ is a linear map from $E$ to itself, where $E$ is some finite dimensional space.
Let $n = \dim(E)$.
Recall that, by definition, the rank of $u$ is $r=\dim(u(E))$.
Suppose that $r=1$. Then $\dim(\ker(u))=n-1$.
Since the multiplicity of an eigenvalue as at least the dimension of the corresponding eigenspace, we get that $0$ is an eigenvalue with multiplicity at least $n-1$.
And since the sum of all eigenvalues (counted with multiplicity) is $\mathrm{tr}\,(u)$, the last eigenvalue is $\mathrm{tr}\,(u)$.
You can now, given some square matrix $A$, apply this to the map canonically attached to $A$.