Prove if $\dim(\ker(T))=n-1$ then $T$ has at most 2 distinct eigenvalues

Question

Let $V$ vector space with $\dim(V)=n$, and let $T: V \rightarrow V$ a linear operator such that $$\dim(\ker(T))=n-1$$

Prove: $T$ has at most 2 distinct eigenvalues.

I know that $\dim(\mathrm{im}(T))=1$ and it has to do with geometric multiplicty for finding an upper limit, but I don't know.

I'd appreciate a full exact proof in order to learn perfect writing. Thanks!

Answer 1

If $n>1$, then $0$ is an eigenvalue. Let $v$ be a generator of the image $T(v)=cv$. Let $w$ another eigenvalue. Write $w=u+av, u\in ker(T) $, $T(w) =aT(v) $ implies that $w=av$.

Answer 2

If you denote $E_\lambda$ the eigenspace relative to the eigenvalue $\lambda$ the simple observation is that $$ \ker(T)=E_0 $$ Since eigenspaces are linearly independent, the hypothesis $\dim(\ker(T))=n-1$ leaves space for at most one more $1$-dimensional eigenspace, hence for at most one eigenvalue $\lambda\neq0$.

Answer 3

If $n=1$, then $T$ has exactly one eigenvalue, since $T(v)=av$ for some scalar $a$.

Otherwise, $\ker T\neq\{0\}$ and therefore $0$ is an eigenvalue. If $T$ has two other distinct eigenvalues $\lambda_1$ and $\lambda_2$, let $v_1$ and $v_2$ be eigenvectors whose eigenvalues are $\lambda_1$ and $\lambda_2$ respectively. Then, $T(v_1)=\lambda_1v_1$ and $T(v_2)=\lambda_2v_2$ and therefore $v_1,v_2\in\operatorname{Im}T$. So, $\dim\operatorname{Im}T\geqslant 2$ and, by the rank-nullity theorem, $\dim\ker T\leqslant n-2$.