LINEAR-ALGEBRA · Interactive Practice | Unit 23 · Video 4

From Dot Products to Fourier: Orthonormal Bases

IKey Formulas

Formula	Name	Description
$q_i^T q_j = \delta_{ij}$	Orthonormality	$1$ if $i=j$ , else $0$
$x_k = q_k^T v$	Coordinate extraction	Dot product with basis vector
$Q^T Q = I \Rightarrow Q^{-1} = Q^T$	Inverse trick	True for square orthonormal $Q$
$\langle f, g \rangle = \int_0^{2\pi} f(x)g(x)\, dx$	Function inner product	Replace sum with integral
$a_k = \frac{1}{\pi}\int_0^{2\pi} f(x)\cos(kx)\, dx$	Fourier coefficient	Project $f$ onto $\cos(kx)$

IIVisualization 1 — Coordinates as Dot Products

For an orthonormal basis, each coordinate is a single dot product: $x_k = q_k^{T} v$ .

Basis angle θ 30°

💡 Challenge: drag $v$ , then rotate the basis until $q_1$ lines up with it — now $x_2 = 0$ and $v$ collapses to a single coordinate.

IIIVisualization 2 — Orthogonality of Sine and Cosine

Distinct Fourier modes are orthogonal: the positive and negative areas of their product cancel to zero.

f(x) g(x)

💡 This is the same orthogonality that gave $q_i^{T} q_j = 0$ for vectors — now lifted from finite sums to integrals over functions.

IVVisualization 3 — Building a Square Wave

Add orthogonal sine modes one at a time and a jagged square wave emerges from smooth curves.

Terms 1

💡 Each mode is a projection onto one orthogonal direction — the same trick as $x_k = q_k^{T} v$ , now giving $b_k = \frac{1}{\pi}\int_0^{2\pi} f(x)\sin(kx)\,dx$ .

VQuiz Questions

Question 1

Let $\{q_1, q_2\}$ be an orthonormal basis with $q_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , $q_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , and let $v = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ .

What is the coordinate $x_1 = q_1^T v$ ?

✅ Correct! $x_1 = q_1^T v = 3$ .

❌ Not quite. Compute the dot product $q_1^T v$ entry by entry.

Show solution

Solution:

The trick for orthonormal bases is $x_k = q_k^T v$ .

$x_1 = q_1^T v = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ -4 \end{pmatrix} = (1)(3) + (0)(-4) = 3.$

No row reduction needed — just a dot product.

Question 2

True or False: If $Q$ is a square matrix whose columns form an orthonormal basis, then $Q^{-1} = Q^T$ .

✅ Correct! Square + orthonormal columns ⟹ $Q^{-1} = Q^T$ .

❌ Not quite. Re-read the squareness condition — it makes $Q^T Q = I$ enough.

Show solution

Solution:

Stacking the orthonormality conditions $q_i^T q_j = \delta_{ij}$ as a matrix product gives $Q^T Q = I.$

For a square matrix, this single identity is enough to conclude $Q^{-1} = Q^T$ (both products $Q^T Q$ and $Q Q^T$ equal $I$ ).

For a non-square $Q$ with orthonormal columns, $Q^T Q = I$ still holds but $Q Q^T \neq I$ in general, so $Q$ has no two-sided inverse. The question specifies square $Q$ , so the statement is True.

Question 3

Consider the inner product $\displaystyle \langle f, g \rangle = \int_0^{2\pi} f(x)\, g(x)\, dx$ .

Which of the following pairs is NOT orthogonal under this inner product?

Select the pair that is NOT orthogonal:

✅ Correct! $\int_0^{2\pi}\sin^2(x)\,dx = \pi$ , the squared norm of $\sin(x)$ .

❌ Not quite. Three of these pairs integrate to zero — find the one that doesn't.

Show solution

Solution:

Two functions are orthogonal iff their inner product is $0$ .

$\int_0^{2\pi} \sin(x)\cos(x)\, dx = \tfrac{1}{2}\int_0^{2\pi}\sin(2x)\,dx = 0$ . Orthogonal.
$\int_0^{2\pi} \cos(x)\cos(2x)\, dx = 0$ (Fourier basis). Orthogonal.
$\int_0^{2\pi} 1 \cdot \cos(x)\, dx = 0$ . Orthogonal.
$\int_0^{2\pi} \sin^2(x)\, dx = \pi \neq 0$ . Not orthogonal — this is a function paired with itself, giving its squared norm.

So $\sin(x)$ and $\sin(x)$ is the non-orthogonal pair.

Question 4

The Fourier coefficient formula is $a_k = \frac{1}{\pi}\int_0^{2\pi} f(x)\cos(kx)\, dx.$

What is the role of the prefactor $\dfrac{1}{\pi}$ ?

✅ Correct! The $1/\pi$ exactly compensates for $\|\cos(kx)\|^2 = \pi$ .

❌ Not quite. Think about $\int_0^{2\pi}\cos^2(kx)\,dx$ — it's the squared norm of the basis function.

Show solution

Solution:

When we apply the inner-product trick to $f(x) = a_0 + \sum_k(a_k\cos(kx) + b_k\sin(kx))$ and multiply by $\cos(kx)$ before integrating, every cross term vanishes by orthogonality. The one surviving term is

$\int_0^{2\pi} f(x)\cos(kx)\, dx = a_k \int_0^{2\pi} \cos^2(kx)\, dx = a_k \cdot \pi.$

So $a_k = \tfrac{1}{\pi}\int_0^{2\pi} f(x)\cos(kx)\, dx$ .

The Fourier basis is orthogonal but not orthonormal — each $\cos(kx)$ has squared norm $\pi$ , not $1$ . The $1/\pi$ prefactor is precisely the normalization fix. If we rescaled each basis function by $1/\sqrt{\pi}$ , we'd get a true orthonormal basis and the prefactor would disappear.

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