# Orthogonal projection onto affine subspaces formula

Essentially what I want to understand is the derivation to this formula.

$$π_L(x) = x_0 + π_U (x − x_0)$$,

where $$π_L$$ is the projection mapping to affine space L, $$π_U$$ is the projection mapping to vector space U. $$L = x_0 + U$$

In a 2D/3D space I can visualize how this formula works.

(This image is from the book Mathematics in Machine Learning)

Subtracting $$x_0$$ from $$x$$, I can find $$π_U(x-x_0)$$, the projection of $$x-x_0$$ onto U, and then use it to find the projection of $$x$$ onto L, by adding $$x_0$$ to it.

But I'm struggling to imagine, or derive this formula for vector spaces of more dimensions.

I would appreciate it if someone gave me a derivation (or an idea) for it, or maybe just an intuitive way to understand this concept.

You can think of the term $$x-x_0$$ as moving the origin to $$x_0$$. Then the algorithm can be described in three steps. Move the origin to $$x_0$$ so that the plane goes through the origin, calculate the linear orthogonal projection onto the plane, and finally move the origin back to $$0$$.
These steps are applied right to left in the formula. First, calculate $$x_0-x$$ to move the origin, then project onto the now linear subspace with $$\pi_U(x-x_0)$$ and finally move the origin back with $$x_0 + \pi_U(x-x_0)$$. Moving the origin back and forth without the projection would be the algorithm $$x_0 + (x - x_0)=x$$. In general in an affine space you can linearize it by choosing an origin $$x_0$$ and using $$x-x_0$$ as vectors since $$x_0-x_0=0$$.
• @KraZZ you can prove that an affine space can be made into a vector space with $x_0$ as the origin by describing all the vectors as $v=x-x_0$ and verifying the axioms. This is relatively straightforward. If you're familiar with the proof of how to orthogonally project onto a linear subspace then you can likely prove translations are isometries as well. That's sufficient for the algorithm to work given the orthogonal projection. Jan 25, 2021 at 16:58