Projection onto a plane I was looking at this post ($3D$ projection onto a plane) in which the answer describes how to project a given set of points onto any arbitrary plane.  However, this transformation is still of the form $\mathbb{R}^3 \to \mathbb{R}^3$ and I would like to obtain a transformation of the form $\mathbb{R}^3 \to \mathbb{R}^2$.  Essentially, I would like to have a 2-D graph of the points in my plane. How can I do this? 

Suppose that the plane is the column space of a given matrix $\mathrm A \in \mathbb R^{3 \times 2}$
$$\{ \mathrm A \mathrm x \mid \mathrm x \in \mathbb R^2 \} \subset \mathbb R^3$$
Given $\mathrm y \in \mathbb R^3$, we have a linear system $\mathrm A \mathrm x = \mathrm y$. In general, $\mathrm y$ is not in the column space of $\mathrm A$. Hence, we project $\mathrm y$ onto the column space of $\mathrm A$. We first left-multiply both sides by $\mathrm A^T$, which produces $\mathrm A^T \mathrm A \mathrm x = \mathrm A^T \mathrm y$ (which always has a solution). As $\mathrm A$ has full column rank (otherwise its column space would be a line, not a plane), then $\mathrm A^T \mathrm A$ is invertible. Hence,
$$\hat{\mathrm x} := (\mathrm A^T \mathrm A)^{-1 } \mathrm A^T \mathrm y$$
gives us the coordinates of the projection of $\mathrm y$ onto the plane, using the basis formed by the two linearly independent columns of $\mathrm A$. Thus, the projection is
$$\mathrm y \mapsto (\mathrm A^T \mathrm A)^{-1 } \mathrm A^T \mathrm y$$
If the columns of $\mathrm A$ are orthonormal, then $\mathrm A^T \mathrm A = \mathrm I_2$ and the projection is simply $\mathrm y \mapsto \mathrm A^T \mathrm y$.
