Endomorphism with all eigenvectors

Let $u$ be an endomorphism of a vector space $E$ such that, for all $x$ in $E\backslash \{0\}$ , $x$ is an eigenvector of $u$ .

What is $u$ ?

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Solution by Anduril

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For every non zero x there is $\lambda_{x}$ (unique) so that $f(x)=\lambda_{x}x$ . Let’s show that this scalar is independant of x.
Take x and y.
If they are proportionate the result is obvious.
Assume they are not proportionate : $f(x)+f(y)=f(x+y)=\lambda_{x+y}(x+y)$
Hence, $\lambda_{y}y=(\lambda_{x+y}-\lambda_{x})x+\lambda_{x+y}y$
(difference of the two preceding relations).

Finally, we conclude using the liberty of the (x,y) family.

So the solutions are the $\lambda Id$ , with $\lambda$ scalars.

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