Finding image and kernel of a linear map

Question

Let $\bf{n}$ be a unit vector in $\mathbb{R}^3$. Find the image and kernel of the linear map $\tau:\bf{x}\rightarrow\bf{x'}= \bf{x}-(\bf{x}\cdot\bf{n})\bf{n}$.

My attempt: The kernel is the set of values of $u$ for which $\tau(u)=0$. As such we need $u-(u\cdot n)n=0$, that is $u=(u\cdot n)n$. Dot product on the LHS: $u\cdot u=u\cdot (u\cdot n)n$. This gives $|u|^2=0$ so only the origin satisfies this. I've no idea about the image - how would one even go about finding it?

Answer 1

You have$$\tau(\mathbf x)=0\iff \mathbf x=(\mathbf x.\mathbf n)\mathbf n\implies\mathbf x=\lambda\mathbf n,$$for some $\lambda\in\Bbb R$. And if $\mathbf x=\lambda\mathbf n$ for some $\lambda\in\Bbb R$, then $\tau(\mathbf x)=\lambda\tau(\mathbf n)=0$. So $\ker\tau=\Bbb R\mathbf n$.

If $\mathbf x.\mathbf n=0$, then $\tau(\mathbf x)=\mathbf x$. It's now easy to see that $\tau\left(\Bbb R^3\right)=\mathbf n^\perp$.

Answer 2

The simplest way is to recognize that $\tau$ is the orthogonal projection onto $\text{span}(n)^{\perp}$. Thus the image is $\text{span}(n)^{\perp}$ and the kernel is $\text{span}(n)$.