Suppose I have the hyperplane in $\mathbb{R}^p$ described by the equation $$\beta_1 x_1 + \cdots + \beta_p x_p = 0.$$ This hyperplane can also be thought of as the orthogonal complement of the set containing the vector $\beta = (\beta_1, \cdots, \beta_p)^T$. There are many ways to obtain a basis of this subspace, but one way is using the following scheme:
$$v_1 = \begin{pmatrix} 1 \\ -\beta_1/\beta_2 \\ 0 \\ 0 \\ \vdots \\ 0\end{pmatrix},\ v_2 = \begin{pmatrix} 0 \\ 1 \\ -\beta_2/\beta_3 \\ 0 \\ \vdots \\ 0\end{pmatrix}, \cdots$$
If we arrange these vectors into a matrix $A$, we wish to calculate the projection $A(A^T A)^{-1} A^T$. But this doesn't seem so easy.
Is there a simpler way of doing this that I'm overlooking?
Without loss on generality assume that $\|\beta\|^2=1.$ Then $x\mapsto p(x)=x-\langle x,\beta\rangle \beta$ is in $H$ and $x-p(x)$ is orthogonal to $H.$ The representative matrix of $p$ in the canonical basis of $R^n$ is $I_n-\beta\beta^T.$