Projection of a hyperplane

Question

Let $\alpha$ be a vector in $X$ such that $||\alpha||=1$ and let $\beta\in\mathbb{R}$. Consider the hyperplane $$C= x\in X:\langle a,x\rangle=\beta.$$ Prove that $$P_c(x)=x-(\langle a,x \rangle-\beta)\alpha.$$ I am hoping someone can help me make some progress on this. I have three immediate questions:

1. how should I approach this problem

2. if $\langle a,x\rangle=\beta$ then shouldn't $\langle a,x\rangle-\beta=0$

3. how does $||\alpha||=1$ come into this

Answer 1

By definition, we have \begin{equation} P_C(x) := \underset{z \in C}{\text{argmin }}\frac{1}{2}\|z-x\|^2 \end{equation}

(i.e, we seek the closest point to $x$ which lies in on the hyperplane $C := \{z \in X | \langle \alpha, z \rangle = \beta\}$.

Hint: To solve the above problem, use the Method of Lagrange multipliers.

The functional to minimize is $L(z, \lambda) = \frac{1}{2}\|z-x\|^2 + \lambda (\langle \alpha, z\rangle - \beta)$, where $\lambda \in \mathbb{R}$ is a Langrange multiplier to be determined. Differentiating w.r.t to $z$ and setting to $0$ yields $z - x + \lambda\alpha = 0$, i.e $z = x - \lambda \alpha$. Imposing the constraint $z \in C$ yields $\langle \alpha, x - \lambda \alpha\rangle - \beta = 0$, i.e $\|\alpha\|^2\lambda = \langle \alpha, x\rangle - \beta$. Since $\|\alpha\| = 1$ by hypothesis, we get $\lambda = \langle \alpha, x\rangle - \beta$.

Putting things together, we recover the desired solution $P_C(x) = x - \lambda \alpha = x - (\langle \alpha, x\rangle - \beta)\alpha$.

Answer 2

1. Show that if $P_c(x)$ defined by the expression, then $\langle \alpha, P_c(x) \rangle = \beta$ and that $ x - P^c(x) \perp C$.
2. Yes, that is how subtraction works in the real numbers.
3. In the expression for $P_c(x)$. When $\|\alpha\|$ is not unital, that expression has to be different. You could multiply $\alpha$ and $\beta$ by the same constant multiplier and the condition would still hold for the same set $C$. But in many calculations on hyperplanes, it is useful to have the plane normal to be unital. This includes the expression for $P_c(x)$, which would need the 2nd term divided by $\|\alpha\|^2$ otherwise. Also, when $\|\alpha\| = 1, \beta$ is the distance from the origin to the hyperplane.