I've been struggling all day with this question. I tried to come up with a proof which shows that such an endomorphism does NOT exist, but I'm not sure it is correct.
- Let $ B = (b_1, b_2, b_3, b_4)$ be a base of $ \mathbb{R}^4$;
- Since $ \text{dim}(\mathbb{R}^4) = \text{dim}(\text{ker}f) + \text{dim}(\text{im}f)$ , then $\text{dim}(\text{ker}f) = \text{dim}(\text{im}f) = 2$;
- We begin building an endomorphism such that $f(b_1) = 0_{\mathbb{R}^4}$ and $f(b_2) = 0_{\mathbb{R}^4}$ (I don't think it matters which vectors we choose to span the kernel). Consequently, $\text{mg}(0) = 2$;
- If $\text{dim}(\text{im}f) = 2$ and $\text{im}(f) = \text{ker}(f)$, then it must be that $f(b_3) = b_1$ and $f(b_4) = b_2$;
- By doing so, $b_1$ and $b_2$ are the only eigenvectors of $f$, and thus we can't find a base of $\mathbb{R}^4$ made of eigenvectors.
Please bear in mind that my knowledge of linear algebra stops at diagonalization, and that this is my very first attempt at making a proof.
If I'm wrong and building such an endomorphism is actually possible, then what am I missing?
Here is a very quick proof: show that because $\ker (f) = \operatorname{im}(f)$, it must hold that $f \neq 0$ but $f^2 = 0$. However, the only diagonalizable endomorphism $f$ for which $f^2 = 0$ is the zero endomorphism. So, $f$ cannot be diagonalizable.
Regarding the points that you have written: first of all, you have not proved that $b_1,b_2$ are the only eigenvectors of $f$. Second, showing that the example you tried to make failed to be diagonalizable while satisfying the condition does not prove that there are no such endomorphisms.