Diagonalizable endomorphism with kernel equal to image

Question

Is it possible to determine a diagonalizable endomorphism $f: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ such that $\ker f = \text{Im } f$ ? If so, could you give me an example?

Answer 1

No, there is no such endomorphism.

Assume ${\rm im\,}f\subseteq \ker f$ and $f$ is diagonalizable, then $f=0$:

Let $v_i$ be a basis that diagonalizes $f$, i.e. $f(v_i)=\lambda_iv_i$ with the diagonal entries $\lambda_i$.
Then, $\lambda_iv_i\in {\rm im\,}f$, so if $\lambda_i\ne 0$ then $v_i\in {\rm im\,}f\subseteq \ker f$ would imply $f(v_i)=0$, thus $\lambda_i=0$.