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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.

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  • $\begingroup$ I believe you when you say you're lost, but please tell us what you've tried or thought of. Even a sketch is a good start. $\endgroup$
    – nomen
    Commented Jan 14, 2020 at 22:33
  • $\begingroup$ You have no $k$ in your formula $f(e_1)=\dots$. $\endgroup$
    – Bernard
    Commented Jan 14, 2020 at 22:34
  • $\begingroup$ 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. $\endgroup$
    – user729424
    Commented Jan 14, 2020 at 22:51
  • $\begingroup$ You certainly have not typed what you think you typed. Please be very careful. $\endgroup$ Commented Jan 14, 2020 at 22:55
  • $\begingroup$ 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 $\sum\lambda_i=0.$ $\endgroup$ Commented Jan 16, 2020 at 14:07

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