Find the symmetric matrix $P$ which is the projection on $V$

Question

I'm dealing with the same problem as in this question: Find the symmetric matrix $P$ for the subspace generated by $v_1$ and $v_2$

But I do not fully understand the answer given in that question. Could someone give me a hint on how to solve the following problem?

Let $V$ the vector subspace of $\mathbb{R}^{3}$ with $dim(V)=2$, generated by the vector:

\begin{align*} v_{1}=\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}, \ \ \ \ \ \ \ \ \ \ v_{2}=\begin{pmatrix} 1\\ -1\\ -1 \end{pmatrix} \end{align*} Find the real symmetric matrix $P\in \mathbb{R}^{3x3}$ that is the projection over $V$, i.e., $P^{2}=P$ and $Pv \in V$ for every $v \in \mathbb{R}^{3}$

Hint: $v_1$ and $v_2$ generates a plane in $\mathbb{R}^{3}$ with a certain normal vector $n$. $Pv$ must be orthogonal to $n$ for every $v\in \mathbb{R}^{3}$

I know that a normal vector to $v_1$ and $v_2$ is $n=v_1 \times v_2 = (2,2,0)$ but I'm not sure how to proceed with the hint.

Answer 1

If $n$ is a normal vector then $P$ is given by $$Pv=v-{1\over \|n\|^2}\langle v, n\rangle n$$ Indeed, $\langle Pv,n\rangle =0$, hence $Pv\in V$ and $$P^2v=Pv-{1\over \|n\|^2}\langle Pv,n\rangle n=Pv$$

Answer 2

So your projection is a map $P: \mathbb{R}^3\rightarrow V$ that assigns $v\in \mathbb{R}^3$ to its projection on $V$, meaning you need to find a way to write what it does concretely; so, use the hint. $n=(2,2,0)$ is orthogonal to $V$ and as such, $\mathscr{B}=\{v_1,v_2,n\}$ is a basis for $\mathbb{R}^3$. Then, your map takes $$v=\sum_{i=1}^3 c_i e_i, \hspace{2mm} e_i \in \mathscr{B}, c_i \in \mathbb{R}$$ to $$Pv=c_1 v_1+c_2 v_2$$ And now you're done. Just pick your favorite basis, feed it to $P$ and assemble the result in columns. For example, picking $\mathscr{B}$, I get as a matrix $$\begin{bmatrix} 1 &0 & 0 \\0 &1 &0 \\ 0&0&0 \end{bmatrix}$$ but you can also feed it the canonical basis and get something else.