Take $x \in \mathbb{R}^n\backslash\{0\}$ and let $L = \text{span}\{x\}$. Now we consider the linear transformation $$T \colon \mathbb{R}^n \to \mathbb{R}^n,$$ to be given by $T(y) = \text{proj}_{L}y = \dfrac{y \cdot x}{x \cdot x}x.$
We would like to find the image and null space of $T.$ Now I know $$\text{Im}(T) = \{T(y): y \in \mathbb{R}^n \} = \left \{\frac{y \cdot x}{x \cdot x}x : y \in \mathbb{R}^n \right \}.$$ Since $\dfrac{y \cdot x}{x \cdot x}$ is a constant, then $\text{Im}(T) = L?$ My only concern here if $\dfrac{y \cdot x}{x \cdot x}$ can be any real number? Then for $\text{Nul}(T) = \{y\in \mathbb{R}^n: T(y) = 0 \} = \{0\}$ since $\dfrac{y \cdot x}{x \cdot x} = 0$ only if $y = 0?$
It is clear that ${\cal R}T \subset L$ since ${\langle y, x \rangle \over \|x\|^2}$ is a scalar. Furthermore, by considering $T(\lambda x) = \lambda x$, we see that ${\cal R}T = L$.
Note that we can write $Ty = {1 \over \|x\|^2} y^T x x = {1 \over \|x\|^2} x^T y x = {1 \over \|x\|^2} x x^T y$, so we have $T = {1 \over \|x\|^2} x x^T$.
We see that $Ty = 0$ iff $x^T y = 0$ iff $y \in L^\bot$. Hence $\ker T = L^\bot$.