I stumbled upon this problem in a list of homework extra exercises. Let $P: \mathbb{R}^{3} \to \mathbb{R}^{3}$ be the orthogonal projection operator of $\mathbb{R}^{3}$ over the plane $x_{1} +2x_{2}-2x_{3}=0$. Find the matrix $[P]_{\beta}$, where $\beta$ is the canonical $\mathbb{R}^{3}$ basis. Given any basis $\beta$ of a linear space $V$ I already know how to compute $[T]_{\beta}$ for a linear operator $T: V \to V$. The problem is that I dont know how the orthogonal projection operator $P: \mathbb{R}^{3} \to \mathbb{R}^{3}$ over the plane $x_{1} +2x_{2}-2x_{3}=0$ is defined. I´ve already googled the definition but all the info is confusing and not homogenous. Can anybody please explain what is the the orthogonal projection operator? How is defined? And what it means in my problem? Then I hopefully compute $[P]_{\beta}$.
Edited: I already have computed $[P]_{\beta}$ obtaining:
$$[P]_{\beta}= \begin{pmatrix}\frac{8}{9} & \frac{6}{5} & \frac{-2}{9} \\ \frac{-2}{9} & \frac{1}{5} & \frac{4}{9}\\ \frac{2}{9} & \frac{4}{5} & \frac{5}{9}\end{pmatrix}$$
Recall that $\Bbb R^3$ can be written as the direct sum of one of its subspaces $S$ and the orthogonal complement of $S$ denoted by $S^\perp$. That is to say, every vector $\mathbf x\in\Bbb R^3$ can be written uniquely as the sum of some $u\in S,v\in S^\perp$. The projection operator of $S$ precisely takes $\mathbf x$ and returns the unique $u$ which is the component of $\mathbf x$ lying in $S$.
As an example, let us solve your question. Here $S$ is the plane $x_1+2x_2-2x_3=0$. A basis of $S$ is $\{(-2,1,0),(0,1,1)\}$. Thus $P(-2,1,0)=(-2,1,0)$ and $P(0,1,1)=(0,1,1)$.
Moreover $S^\perp=\{k(1,2,-2):k\in\Bbb R\}$. Thus we also have $P(1,2,-2)=\bf 0$.
Using these three equations and simple addition/subtraction, can you find $Pe_i,e_i\in\beta$, which is the $i^\text{th}$ column of $P$?