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

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?

• You have no $k$ in your formula $f(e_1)=\dots$. Commented Jan 14, 2020 at 22:34
• I'm not sure I believe what we're supposed to prove. Suppose $\sum_{i=0}^n\lambda_i e_i=0$. Then we have that $f(e_1)=\ldots=f(e_n)=\sum_{i=0}^n\lambda_i e_i=0$. Then $f(x)=0$ for all $x$. Hence $f$ is diagonalizable. Commented Jan 14, 2020 at 22:51
• The statement is not true, it lacks the condition that at least one of the $\lambda_i$ is not $0$. If $\lambda_1 = \lambda_2 = \ldots = \lambda_n = 0,$ then $f$ is diagonalizable even if $\sum\lambda_i=0.$ Commented Jan 16, 2020 at 14:07