Linear algebraConcept reference

Projection onto a line

The orthogonal projection of a vector onto the span of a nonzero vector is the multiple whose difference from the original vector is perpendicular to that span.

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Overview
Projection finds the part of a vector that points along a chosen line
Write p=cu
The coefficient can be negative or zero
This is the geometry behind fitting a one-feature linear model with no intercept
Key takeaway
Further questions
Concept connections

Figure 1With b=(2,3) and u=(1,0), the line span(u) is horizontal through O. Its orthogonal projection is p=(2,0). The dashed gap from p to b is perpendicular to the line. Equal scale is 30 pixels per coordinate unit; b is longer than p.

01Projection finds the part of a vector that points along a chosen line#

Projection finds the part of a vector that points along a chosen line. Imagine fitting a target vector using only one allowed direction. You may slide along that line, but cannot move sideways; projection chooses its closest point to the target.

The line span(u) passes through the origin. Its closest point to the tip of b defines the projected vector p. The gap from p to b is perpendicular to the line.

Use $b=(3,1)$ and $u=(1,1)$ . Their dot is 4; u·u=2. The coefficient is 4/2=2, so p=2u=(2,2).

Figure 2For b=(3,1),u=(1,1), the projection is p=(2,2). The line through O in direction u contains p; the dashed vector from p to b is (1,-1), perpendicular to u. Equal scale is 32 pixels per coordinate unit.

For target $b=(2,4)$ and direction $a=(1,1)$ , compute $a\cdot b=6$ and $a\cdot a=2$ . Their ratio 3 is a coefficient. Multiply the entire direction by it to get projected vector $(3,3)$ . The leftover $b-(3,3)=(-1,1)$ has dot product zero with $a$ , checking that no along-line correction remains.

Check your reasoning

$b=(2,6)$ , $u=(1,1)$ . Project b onto span(u).

A(4,4)
B(8,8)
C(-2,2)

Show answer and explanation

(4,4)

The coefficient is 8/2=4.

02Write p=cu#

Write p=cu. A perpendicular gap requires $(b-cu)\cdot u=0$ . Solving for c gives the rule, with $u\ne0$ :

p=\frac{b\cdot u}{u\cdot u}\,u

Check your reasoning

$b=(3,1)$ , $u=(2,0)$ . Find c in the projection p=cu.

Show answer and explanation

1.5

c=6/4=1.5.

03The coefficient can be negative or zero#

The coefficient can be negative or zero. If b·u=0, then p=0. A projection can shorten a vector; it keeps b unchanged only when b already lies on the line. The zero vector cannot supply a line direction.

Check your reasoning

A rail allows multiples of $u=(0,2)$ . Desired $b=(1,-3)$ . Closest allowed move?

A(1,-3)
B(0,-3)
C(0,-12)

Show answer and explanation

(0,-3)

The coefficient is −6/4=−1.5.

04This is the geometry behind fitting a one-feature linear model with no intercept#

This is the geometry behind fitting a one-feature linear model with no intercept. The coefficient says how much of the allowed pattern to use; the projected vector contains the fitted outputs. Those two answers have different shapes and should not be interchanged.

Key takeaway

Compute the coefficient, then scale the nonzero direction.

Compute a complete projection vector onto a nonzero direction’s span

Further questions

Why divide by u·u?

The spanning vector need not be a unit vector. Dividing by its squared length makes the perpendicular-gap condition hold. For u=0 the denominator vanishes and u does not span a line.

01Projection finds the part of a vector that points along a chosen line#

02Write p=cu#

03The coefficient can be negative or zero#

04This is the geometry behind fitting a one-feature linear model with no intercept#

Compute the coefficient, then scale the nonzero direction.

Further questions

Reference this concept