If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues 
Let $T:V\to V$ be a linear operator. If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues.

I have been working on this proof for a few days and I am not sure what direction to really go with it? I feel like starting with the rank nullity theorem is correct and relating that to the sum of eigenspaces may be my next move. Though I cant think of how to bring these two ideas together to create a fluid proof? Thank you for your help...

Hint: Yes, rank nullity theorem is the way to go.
Hint: The reason for the extra $+1$, is because 0 is a possible eigenvalue.
