Okay so I have a vector $\textbf{v}=\begin{bmatrix}1\\-1\\1\end{bmatrix} $. I want to calculate the orthogonal projection of the vector $\textbf{u}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ onto $\textbf{v}$, find the standard matrix of it, give a geometrical argument for whether it is injective/surjective and give a geometrical argument to determine the kernel, null space, image and column space of that standard matrix.
So what I did first was that I calculated the projection using that
$$P_\textbf{v}(\textbf{u}) = \frac{\langle \textbf{v},\textbf{u} \rangle}{\langle\textbf{v},\textbf{v} \rangle}\textbf{v}=\frac{[1,-1,1]\cdot[1,0,0]}{[1,-1,1]\cdot[1,-1,1]}[1,-1,1]=[\frac{1}{3},-\frac{1}{3},\frac{1}{3}]$$
Which is simple enough, but I don't really know how to use this to create a standard matrix. Also I have no idea how to reason geometrically for the questions asked further, so if someone could push me in the right direction, that would be great!
Best Answer
For any linear map $\varphi:\Bbb R^n\to \Bbb R^m$, its standard matrix $M$ can be obtained by putting $\varphi(e_i)$ in the $i$th column, where $(e_1,\dots, e_n)$ is the standard basis of $\Bbb R^n$.
Note that this also means that $\varphi(v)=M\cdot v$ holds for each basis vector, and hence, by taking linear combinations, it must hold for all vectors $v\in\Bbb R^n$.
In the case of the orthogonal projection to a vector $v$, we will obtain $P_v=Q:=\frac{vv^T}{\|v\|^2}$, since it satisfies $Qv=v$ and $Qw=0$ if $w\perp v$.
For the geometrical arguments, draw what $P_v$ being an orthogonal projection means.
Its range (=image=column space) is just the line of $v$ (consisting only of scalar multiples of $v$), which is one dimensional, so the rank is $1$, and it's not surjective.
Its kernel (=null space) consists of exactly the vectors perpendicular to $v$. In an $n$ dimensional space, the orthogonal subspace of a nonzero vector is $n-1$ dimensional, now it's $2$, and it also means that it's not injective, as there are (a whole plane of) nonzero vectors that are mapped to $0$.