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Math · Linear Algebra

Chapter 8: Orthogonality and Applications

Projections

The shadow one vector casts on another.

Lesson
2
Time
About 23 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

The projection of u onto v is the part of u that lies along v. Everything perpendicular is discarded.

The shadow picture

Shine a light straight down onto the line through v. The shadow of u on that line is the projection.

The formula

The projection is (u · v divided by v · v) times v. The fraction is how many copies of v to take.

The remainder

Subtracting the projection from u leaves a vector orthogonal to v. Any vector splits into these two pieces.

When v is a unit vector

If v has length 1, the denominator is 1 and the projection is simply (u · v) times v.

Why it matters

Projection is how you find the closest point in a subspace to a given vector. That is the whole idea behind least squares.

The shadow one vector casts

The projection of u onto v is the part of u lying along v. It is the closest point to u on the line through v, which is the property that makes projection useful.

The formula and its parts

The projection is (u · v / v · v) times v. The dot product measures how much of u lies along v, and dividing by v · v normalises for the length of v.

The remainder is perpendicular

Subtracting the projection from u leaves a vector orthogonal to v. Splitting a vector into parallel and perpendicular parts is a standard move in physics and in least squares.

Projecting onto a subspace

The same idea projects onto a plane or higher subspace, giving the closest point in it. That closest-point property is precisely what least squares exploits.

Step 2: Try It Yourself

Tap and try it out.

Imagine the shadow one arrow casts on the other. Turn them perpendicular and the shadow vanishes to nothing.
  • Vector a(4, 3) · length 5
  • Vector b(5, 0) · length 5

Step 3: Watch an Example

One step at a time.

Watch Nia Cast a Shadow

Nia projects u = ⟨4, 3⟩ onto v = ⟨1, 0⟩.

  1. Step 1

    She computes u · v = 4(1) + 3(0) = 4.

Step 4: Your Turn

Practice makes it stick.

The Shadow

Problem 1 of 2

u = ⟨7, 2⟩ projected onto ⟨1, 0⟩. What is the first entry of the projection?

The Vanishing

Problem 2 of 2

u is orthogonal to v. What is the norm of the projection of u onto v?

Cast the Shadow

1 of 8

u = ⟨5, 9⟩ projected onto ⟨1, 0⟩. What is the first entry?

2 of 8

u = ⟨5, 9⟩ projected onto ⟨0, 1⟩. What is the second entry?

3 of 8

u · v = 12 and v · v = 4. How many copies of v does the projection take?

4 of 8

v is a unit vector and u · v = 6. What is the norm of the projection?

5 of 8

Projecting u onto itself. How many copies of u does it take?

6 of 8

u · v = 0. How many copies of v does the projection take?

7 of 8

Match each piece of the decomposition to what it is.

Tap a card on the left to start.

8 of 8

u · v = 20 and v · v = 5. How many copies of v does the projection take?

Step 5: Quick Check

Show what you know.

Question 1 of 2

u · v = 18 and v · v = 6. How many copies of v does the projection take?

Question 2 of 2

What is left after subtracting the projection of u onto v from u?

What You Learned

  • The projection of u onto v is the part of u lying along v, computed as (u · v)/(v · v) times v.
  • What is left over after subtracting it is orthogonal to v.
  • Projection finds the closest point in a subspace, which is the basis of least squares.