LINEAR-ALGEBRA · Unit 14 · Video 1 · Interactive Practice

One Right Angle Is All You Need — The Projection Matrix

IKey Formulas

Formula	Name	Description
$\hat{x} = \dfrac{\mathbf{a}^T\mathbf{b}}{\mathbf{a}^T\mathbf{a}}$	Scalar coefficient	How far along a to walk
$\mathbf{p} = \hat{x}\,\mathbf{a} = \dfrac{\mathbf{a}^T\mathbf{b}}{\mathbf{a}^T\mathbf{a}}\,\mathbf{a}$	Projection vector	Closest point on line to b
$P = \dfrac{\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}}$	Projection matrix	Maps any vector onto line through a
$P^T = P, \quad P^2 = P$	Key properties	Symmetric and idempotent

IIVisualization 1 — Projecting onto a Line

The projection $\mathbf{p}$ is the closest point on the line through $\mathbf{a}$ to $\mathbf{b}$ ; the error $\mathbf{e} = \mathbf{b} - \mathbf{p}$ always meets the line at a right angle.

💡 Challenge: drag the tip of $\mathbf{b}$ until it lands on the line — the error collapses to zero and $\mathbf{p} = \mathbf{b}$ .

IIIVisualization 2 — Scaling a and b

Scaling $\mathbf{b}$ stretches its projection by the same factor; scaling $\mathbf{a}$ leaves the projection exactly where it was.

Scale factor k 1.00

💡 $P = \mathbf{a}\mathbf{a}^T / \mathbf{a}^T\mathbf{a}$ depends only on the direction of $\mathbf{a}$ : the extra factors of $k$ in numerator and denominator cancel.

IVVisualization 3 — What P Does to Every Vector

The matrix $P$ crushes the whole plane onto the line through $\mathbf{a}$ (rank 1), and applying it a second time changes nothing ( $P^2 = P$ ).

Line angle θ 60°

Pv — every image lands on the line (rank 1)

P²v = Pv — the rings sit on the arrow tips (idempotent)

VQuiz Questions

Question 1

Given $\mathbf{a} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ , what is the scalar coefficient $\hat{x} = \dfrac{\mathbf{a}^T\mathbf{b}}{\mathbf{a}^T\mathbf{a}}$ ?

✅ Correct! aᵀb = 4 and aᵀa = 14, so x̂ = 4/14 = 2/7.

❌ Not quite. Compute aᵀb (numerator) and aᵀa (denominator) separately, then simplify the fraction.

Show solution

Solution:

Compute the dot products step by step:

$\mathbf{a}^T\mathbf{b} = (1)(2) + (2)(1) + (3)(0) = 2 + 2 + 0 = 4$

$\mathbf{a}^T\mathbf{a} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$

$\hat{x} = \frac{4}{14} = \frac{2}{7}$

The answer is 2/7.

Question 2

In the projection matrix formula $P = \dfrac{\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}}$ , the numerator $\mathbf{a}\mathbf{a}^T$ and denominator $\mathbf{a}^T\mathbf{a}$ are:

✅ Correct! Column × row = matrix (outer product); row × column = scalar (inner product). The order matters!

❌ Not quite. Think about dimensions: a is n×1 and aᵀ is 1×n. What shape does each product give?

Show solution

Solution:

$\mathbf{a}\mathbf{a}^T$ is column × row (outer product): $(n \times 1)(1 \times n) = n \times n$ matrix. Every column is a scalar multiple of $\mathbf{a}$ — it's rank 1.
$\mathbf{a}^T\mathbf{a}$ is row × column (inner product / dot product): $(1 \times n)(n \times 1) = 1 \times 1$ scalar.

Key: The order of multiplication matters! $\mathbf{a}\mathbf{a}^T$ (outer) $\neq$ $\mathbf{a}^T\mathbf{a}$ (inner).

Question 3

True or False: Replacing $\mathbf{a}$ with $5\mathbf{a}$ changes the projection matrix $P = \dfrac{\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}}$ .

✅ Correct! The scale factors cancel — P depends only on the direction of a, not its length.

❌ Not quite. Try substituting 5a into the formula and see what cancels in the numerator and denominator.

Show solution

Solution: False.

Substitute $5\mathbf{a}$ for $\mathbf{a}$ :

$P' = \frac{(5\mathbf{a})(5\mathbf{a})^T}{(5\mathbf{a})^T(5\mathbf{a})} = \frac{25\,\mathbf{a}\mathbf{a}^T}{25\,\mathbf{a}^T\mathbf{a}} = \frac{\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}} = P$

The factors of $25$ cancel completely. $P$ depends only on the direction of $\mathbf{a}$ , not its magnitude.

Question 4

You project $\mathbf{b}$ onto the line through $\mathbf{a}$ and get $\mathbf{p} = P\mathbf{b} = [2,\; 4,\; 6]^T$ . Now you project $\mathbf{p}$ onto the same line again. What is $P\mathbf{p}$ ?

✅ Correct! By idempotency (P² = P), projecting again gives the same result. The shadow of a shadow is the shadow itself!

❌ Not quite. Remember P² = P (idempotency). What happens when you project a vector that's already on the line?

Show solution

Solution:

By idempotency ( $P^2 = P$ ):

$P\mathbf{p} = P(P\mathbf{b}) = P^2\mathbf{b} = P\mathbf{b} = \mathbf{p} = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

Geometrically, $\mathbf{p}$ already lies on the line through $\mathbf{a}$ . Projecting a point that's already on the line onto the same line doesn't move it.

The shadow of a shadow is the shadow itself.

Solved: 0 / 4