Q. From a linear operator $f \in L(R4)$ it is known that $f(2, 1, 0, 0) = (0, 0, 1, 2), f(1, −1, 0, 0) = (0, 0, 3, 1)$, and that $\ker f = im f$. Calculate the matrix $F$ of the operator $f$ in the canonical basis.
I tried going with $f(2\vec{e_1})+f(\vec{e_2})=\vec{e_3}+2\vec{e_4}$ and $f(\vec{e_1})-f(\vec{e_2})=3\vec{e_3}+\vec{e_4}$ with $\vec{e_1},\vec{e_2}, \vec{e_3}, \vec{e_4}$ being the canonical basis, with $\vec{e_1}=(1,0,0,0)$, etc. So that we have
$$f(\vec{e_1})=\frac{4}{3}\vec{e_3}+\vec{e_4}$$ and $$f(\vec{e_2})=-\frac{5}{3}\vec{e_3}$$ but from here my teacher uses that $\ker f=im f$ and gets to $$f(\vec{e_3})=f(\vec{e_4})=0$$ which really confuses me. How and why is that equivalent?