I've been trying to solve this problem but I'm missing something:
Derive an endomorfism $f : \mathbb{R}^3 \to \mathbb{R}^3$ such that $\ker(f) : x-2y-z=0$ and $f$ is not diagonalizable.
As the dimension of the subspace associated with $\lambda=0$ is $2$, the algebraic multiplicity of lambda could be $2$ or $3$. It cannot be $2$ because that would imply that there exists another eigenvalue with algebraic multiplicity 1, whose associated subspace would have dimension $1$, and thus $f$ would be diagonalizable. But then I dunno how to choose a vector that is a basis for the range and also keeping that $\lambda=0$ triple. Any help, please?