PROJECTION ON A LINE

Alicia Y. Tsai; Authors: Laurent EI Ghaoui; Contributors: Phan Hoang, Tony Tin, Ha To Tam, Nguyen Van Kep, Le Dinh Nam, Juliette Decugis, Nguyen Quoc Cuong.; Giuseppe C. Calafiore

3 PROJECTION ON A LINE

3.1. Definition

Consider the line in $\mathbb{R}^n$ passing through $x_0 \in \mathbb{R}^n$ and with direction $u \in \mathbb{R}^n$ :

$\begin{align*} \{x_0 + tu: t \in \mathbb{R}\} \end{align*}$

Example 1: A line in $\mathbb{R}^2$ passing through the point $x_0 = (0,1)$ , with direction $u=(0.8944, 0.4472)$ .

The projection of a given point $x$ on the line is a vector $z$ located on the line, that is closest to $x$ (in Euclidean norm). This corresponds to a simple optimization problem:

$\begin{align*} \min\limits_{t} ||x-(x_0+tu)||_2 \end{align*}$

This particular problem is part of a general class of optimization problems known as least-squares. It is also a special case of a Euclidean projection on a general set.


	Example 2: Projection of the vector $x=(1.6, 2.28)$ on a line passing through the origin ( $x_0 = 0$ ) and with (normalized) direction $u = (0.8944, 0.4472)$ .
	At optimality the ‘‘residual’’ vector $x-z$ is orthogonal to the line, hence $z = tu$ , with $t = x^Tu = 2.0035$ . Any other point on the line is farther away from the point $x$ than its projection $z$ is.
	The scalar $t = u^T x$ , ie the scalar product between $x$ and $u$ , is the component of $x$ along the normalized direction $u$ .

3.2. Closed-form expression

Assuming that $u$ is normalized, so that $||u||_2=1$ , the objective function of the projection problem reads, after squaring:

$\begin{align*} ||x-x_0-tu||_2^2 = t^2 - 2tu^T(x-x_0) + ||x-x_0||_2^2 = (t-u^T(x-x_0))^2 + \text{{constant}} \end{align*}$

Thus, the optimal solution to the projection problem is

$\begin{align*} t^* = u^T(x-x_0) \end{align*}$

and the expression for the projected vector is

$\begin{align*} z^* = x_0 + t^*u = x_0 + u^T(x-x_0)u \end{align*}$

The scalar product $u^T(x-x_0)$ is the component of $x-x_0$ along $u$ .

In the case when $u$ is not normalized, the expression is obtained by replacing $u$ with its scaled version $u/||u||_2$ :

$\begin{align*} z^* = x_0 + t^*u = x_0 + \frac{u^T(x-x_0)}{u^T u}u \end{align*}$

3.3. Interpreting the scalar product

We can now interpret the scalar product between two non-zero vectors $x, u$ , by applying the previous derivation to the projection of $x$ on the line ${\bf L}$ of direction $u$ passing through the origin. If $u$ is normalized ( $||u||_2 = 1$ ), then the projection of $x$ on ${\bf L}$ is $z^* = (u^T x)u$ . Its length is $||z^*||_2 = |u^Tx|$ . (See above figure from Example 2.)