Prove that $f$ is diagonalizable if and only if $\sum_{i=0}^n \lambda_{i} \neq 0$

Question

I'm a bit lost with this exercise.

Let $e_{1},...,e_{n}$ be a basis of the $E$ $\;K$-vector space and $f \in \operatorname{End}(E)$ such that $f(e_{1}) = ...=f(e_{n}) = \sum_{i=0}^n \lambda_{i}e_{i}$, where $k \in K$ (eigenvalue). Prove that $f$ is diagonalizable if and only if $\sum_{i=0}^n \lambda_{i} \neq 0$.

How is it done?

If $f(e_{1}) = \lambda_{1}e_{1}$,$\;\;\;f(e_{2}) = \lambda_{1}e_{1} + \lambda_{2}e_{2}$ and so on.., then $\lambda_{1}$ must be the same as $\lambda_{2}$, so $f(e_{2}) = \lambda(e_{1} + e_{2})$, right?

Thanks in advance.