Rank of a linear map

Matrix.
$$A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 4 & 0 \\ 3 & 6 & 1 \end{pmatrix}.$$

Row reduction. $R_2\leftarrow R_2-2R_1$ gives a zero row, and $R_3\leftarrow R_3-3R_1$ gives
$$\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
There are two nonzero rows, so $\operatorname{rank}(f)=2$.

Kernel. $Ax=0$ means $x+2y=0$ and $3x+6y+z=0$, hence $z=0$ and $x=-2y$: $\ker f=\operatorname{span}\big((-2,1,0)\big)$.

Conclusion. $\operatorname{rank}(f)=2$ and $\dim\ker f=1$, so $2+1=3=\dim\mathbb{R}^3$.