How to find a projection matrix for a line

Projecting a vector onto a line keeps the part that points along the line. We can calculate that part using a dot product. Then we can write the whole operation as a matrix that works for every input vector.

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Use a nonzero vector along the line

If $u$ points along the line, the projection of $v$ is $\frac{v·u}{u·u}u$. The denominator divides by the squared length of $u$, so the formula also works when $u$ is longer or shorter than a unit vector.

Project a vector onto the line through one, three

Let $u=(1,3)$ and $v=(0,2)$. The two dot products are $v·u=6$ and $u·u=10$.

$\frac{6}{10}(1,3)=\left(\frac{3}{5},\frac{9}{5}\right)$

The dashed segment is perpendicular to the line. The projection vector ends at the foot of that segment.

The projection is $(0.6,1.8)$. The remaining part is $v-(0.6,1.8)=(-0.6,0.2)$, whose dot product with $(1,3)$ is $0$.

Build a matrix that gives the same projection

For an input $(x,y)$, the projection formula gives $\frac{x+3y}{10}(1,3)$. Its components are $\frac{x+3y}{10}$ and $\frac{3x+9y}{10}$, so the projection matrix is

$P=\left[\begin{array}{rr}\frac{1}{10}& \frac{3}{10}\\ \frac{3}{10}& \frac{9}{10}\end{array}\right]$

Once a vector lies on the line, projecting it again leaves it in place. In matrix form, $P^2=P$.

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