Basis and kernel of linear transformation

Question

Let $S : \mathbb{R^3} \rightarrow \mathbb{R^3}$ defined as $S(x) = (\frac{\langle u,x\rangle}{\langle u,u\rangle})u$ with $u := [u_1,u_2,u_3]$ a vector(not the zero vector) in $\mathbb{R^3}$ . What would be the standard basis for S and what would be the dim of the kernel (i assume that $u_3 \neq 0 $.)

Answer 1

Hint: Since $S$ is projection onto $u$, the kernel is everything orthogonal to $u$. So $v\in\operatorname{ker} S\iff v\cdot u=0$. So the kernel is $2$ dimensional: choose anything for $v_1$ and $v_2$, then $v_3=\frac{-u_1v_1-u_2v_2}{u_3}$.

I presume you mean the matrix rel the standard basis $\beta=\{e_1,e_2,e_3\}$. Simply apply $S$ to each $e_i$ and write the result in terms of $\beta$ to get the $i$th column of the matrix. For the first column, I get $\frac1{u_1^2+u_2^2+u_3^2}\cdot (u_1^2,u_1u_2,u_1u_3)$. I leave the other two to you.

If, on the other hand, you wanted a basis for the image of $S$, you could use $\{u\}$.