Let $T : V\to V$ be a linear transformation such that dimension of $\operatorname{Range}(T)= k \leq n$, where the dimension of $V$ is $n$. Show that $T$ can have at most $(k+1)$ distinct eigenvalues.
I can realise that the rank will correspond to the number of non-zero eigenvalues (counted up to multiplicity) and the nullity will correspond to the 0 eigenvalue (counted up to multiplicity) but I cannot design an analytical proof of this.
Thanks for any help .