Linear-Algebra-Done-Right · Unit 15 · Video 1 · Interactive Practice

Orthonormal Bases: Bessel's Inequality and Parseval

IKey Formulas

Formula	Statement	Hypotheses
$\langle e_j, e_k \rangle = 1$ if $j = k$ , $\ \langle e_j, e_k \rangle = 0$ if $j \neq k$	Definition of an orthonormal list: every vector has norm $1$ and every two are orthogonal	$e_1, \dots, e_m$ in an inner product space $V$
$\\|a_1 e_1 + \cdots + a_m e_m\\|^2 = \|a_1\|^2 + \cdots + \|a_m\|^2$	Norm of an orthonormal combination; with the combination equal to $0$ it forces every $a_k = 0$ , so orthonormal lists are linearly independent	$e_1, \dots, e_m$ orthonormal; $a_1, \dots, a_m \in \mathbf{F}$
$\|\langle v, e_1 \rangle\|^2 + \cdots + \|\langle v, e_m \rangle\|^2 \le \\|v\\|^2$	Bessel's inequality; the gap is $\\|w\\|^2$ with $w = v - \sum_{k=1}^{m} \langle v, e_k \rangle e_k$	$e_1, \dots, e_m$ orthonormal; $v \in V$
$v = \sum_{k=1}^{n} \langle v, e_k \rangle e_k$ , $\ \\|v\\|^2 = \sum_{k=1}^{n} \|\langle v, e_k \rangle\|^2$	Coordinates are inner products, and Parseval's identity (Bessel with equality); also $\langle u, v \rangle = \sum_{k=1}^{n} \langle u, e_k \rangle \overline{\langle v, e_k \rangle}$	$e_1, \dots, e_n$ an orthonormal basis of $V$ ; $u, v \in V$

Key Insight: In the orthonormal basis of $\mathbf{F}^4$ with entries $\pm\tfrac12$ , the four coefficients of $v = (1, 2, 4, 7)$ are four separate inner products, $7$ , $-4$ , $1$ , $2$ , with no system to solve, and Parseval checks them: $49 + 16 + 1 + 4 = 70 = \|v\|^2$ .

IIInner Products of Functions Are Signed Areas

Each entry $\langle f, g \rangle = \int fg$ of the table is a signed area; a list is orthonormal exactly when the table is $1$ on the diagonal and $0$ elsewhere.

list

💡 Dividing $1, x, x^2$ by their norms fixes the diagonal, but a nonzero multiple of a nonzero inner product is never zero; the next video builds an orthonormal list from $1, x, x^2$ by Gram–Schmidt.

IIIBessel's Inequality in the Plane

The sum of the squared coefficients along an orthonormal list never exceeds $\|v\|^2$ ; the shortfall is $\|w\|^2$ , the part of $v$ the list misses.

orthonormal list

IVCoordinates in an Orthonormal Basis of F⁴

With an orthonormal basis each coefficient of $v$ is one inner product; for a list with entries $\pm\tfrac12$ that is not orthonormal, $\sum \langle v, e_k \rangle e_k$ usually fails to rebuild $v$ .

VQuiz Questions

Problem 1 · A Coefficient as an Inner Product

Given: the orthonormal basis of $\mathbf{F}^4$ from the video, $e_1 = \left(\tfrac12, \tfrac12, \tfrac12, \tfrac12\right)$ , $e_2 = \left(\tfrac12, \tfrac12, -\tfrac12, -\tfrac12\right)$ , $e_3 = \left(\tfrac12, -\tfrac12, -\tfrac12, \tfrac12\right)$ , $e_4 = \left(-\tfrac12, \tfrac12, -\tfrac12, \tfrac12\right)$ , and $v = (5, 1, 3, -1)$ . Find the coefficient $a_4$ in $v = a_1 e_1 + a_2 e_2 + a_3 e_3 + a_4 e_4$ .

✅ Correct!

a_4 = \langle v, e_4 \rangle = \tfrac12(-5 + 1 - 3 - 1) = -4

, one inner product and no system of equations.

❌ Check the signs. The sign pattern of

e_4

-, +, -, +

, so the signed sum is

-5 + 1 - 3 - 1 = -8

, which is negative.

❌ Close, but…

-8

is the signed sum

-5 + 1 - 3 - 1

; each entry of

e_4

\pm\tfrac12

, not

\pm 1

, so the inner product is half of it.

❌ Not quite.

-2 = \tfrac14(-8)

uses the factor

\tfrac12

twice. The coefficient is

a_4 = \langle v, e_4 \rangle

, with one factor

\tfrac12

from the entries of

e_4

and no division by

\|e_4\|^2 = 1

, so

a_4 = \tfrac12(-8)

❌ Not quite. Use

a_k = \langle v, e_k \rangle

with the entries of

e_4

Show solution

Since $e_1, e_2, e_3, e_4$ is an orthonormal basis, the first part of the theorem gives $a_k = \langle v, e_k \rangle$ . The entries of $e_4$ are $-\tfrac12, \tfrac12, -\tfrac12, \tfrac12$ , so

a_4 = \langle v, e_4 \rangle = \tfrac12\big(-5 + 1 - 3 + (-1)\big) = \tfrac12(-8) = -4.

The other coefficients are $a_1 = \tfrac12(5 + 1 + 3 - 1) = 4$ , $a_2 = \tfrac12(5 + 1 - 3 + 1) = 2$ , $a_3 = \tfrac12(5 - 1 - 3 - 1) = 0$ . Parseval checks the set: $4^2 + 2^2 + 0^2 + (-4)^2 = 36 = 25 + 1 + 9 + 1 = \|v\|^2$ .

Problem 2 · Normalizing Is Not Enough

Given: $\mathcal{P}_2(\mathbf{R})$ with $\langle p, q \rangle = \int_{-1}^{1} pq$ , and the normalized list $\tfrac{1}{\sqrt2},\ \sqrt{\tfrac32}\,x,\ \sqrt{\tfrac52}\,x^2$ . Find $\left\langle \tfrac{1}{\sqrt2},\ \sqrt{\tfrac52}\,x^2 \right\rangle$ .

✅ Correct! The inner product is

\tfrac{\sqrt5}{2} \cdot \tfrac23 = \tfrac{\sqrt5}{3} \neq 0

, so the normalized list is still not orthonormal.

❌ Not quite. Dividing by norms makes each vector a unit vector, but it multiplies

\langle 1, x^2 \rangle = \tfrac23

by a positive number, and a positive multiple of

\tfrac23

is not

0

❌ Close, but…

\tfrac23

\langle 1, x^2 \rangle

before rescaling; the factors

\tfrac{1}{\sqrt2}

and

\sqrt{\tfrac52}

multiply it.

❌ Not quite. Each vector has norm

1

, but that fixes only

\langle e_k, e_k \rangle

; this is an inner product of two different vectors.

❌ Not quite. Pull the constants out and integrate

x^2

over

[-1, 1]

Show solution

Constants come out of the integral:

\left\langle \tfrac{1}{\sqrt2},\ \sqrt{\tfrac52}\,x^2 \right\rangle = \frac{1}{\sqrt2}\sqrt{\frac52} \int_{-1}^{1} x^2\,dx = \frac{\sqrt5}{2} \cdot \frac23 = \frac{\sqrt5}{3} \approx 0.745.

Each vector of the list has norm $1$ (for instance $\int_{-1}^{1} \tfrac52 x^4\,dx = \tfrac52 \cdot \tfrac25 = 1$ ), and $\sqrt{\tfrac32}\,x$ is orthogonal to the other two because $x$ and $x^3$ are odd. But the first and third vectors are not orthogonal, so the list is not orthonormal: scaling cannot turn a nonzero inner product into $0$ .

Problem 3 · The Gap in Bessel's Inequality

Given: the orthonormal list $e_1 = \left(\tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt3}\right)$ , $e_2 = \left(-\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}, 0\right)$ in $\mathbf{R}^3$ and $v = (3, 0, 3)$ . Find $\|w\|^2$ , where $w = v - \langle v, e_1 \rangle e_1 - \langle v, e_2 \rangle e_2$ .

✅ Correct!

\|w\|^2 = \|v\|^2 - |\langle v, e_1 \rangle|^2 - |\langle v, e_2 \rangle|^2 = 18 - 12 - \tfrac92 = \tfrac32

❌ Not quite. Bessel's inequality is an equality only when

v

lies in the span of the list; two vectors span only a plane in

\mathbf{R}^3

, and

v

is not in it.

❌ Close, but…

18 - 12 = 6

removes only the

e_1

part;

|\langle v, e_2 \rangle|^2 = \tfrac92

must be subtracted too.

❌ Close, but…

12 + \tfrac92 = \tfrac{33}{2}

\|u\|^2

, the part of

\|v\|^2

the list captures;

\|w\|^2

is what remains.

❌ Not quite. Use

\|v\|^2 = \|u\|^2 + \|w\|^2

from the proof of Bessel's inequality.

Show solution

From the proof of Bessel's inequality, $\|v\|^2 = \|u\|^2 + \|w\|^2$ with $\|u\|^2 = |\langle v, e_1 \rangle|^2 + |\langle v, e_2 \rangle|^2$ . Compute:

\langle v, e_1 \rangle = \frac{3 + 0 + 3}{\sqrt3} = \frac{6}{\sqrt3},\quad |\langle v, e_1 \rangle|^2 = 12; \qquad \langle v, e_2 \rangle = \frac{-3 + 0 + 0}{\sqrt2},\quad |\langle v, e_2 \rangle|^2 = \frac92.

\|w\|^2 = \|v\|^2 - \|u\|^2 = 18 - 12 - \frac92 = \frac32.

Check directly: $u = 2(1, 1, 1) + \tfrac32(1, -1, 0) = \left(\tfrac72, \tfrac12, 2\right)$ , so $w = v - u = \left(-\tfrac12, -\tfrac12, 1\right)$ and $\|w\|^2 = \tfrac14 + \tfrac14 + 1 = \tfrac32$ . As in the video, $w$ is a multiple of $e_3 = \tfrac{1}{\sqrt6}(1, 1, -2)$ , and $|\langle v, e_3 \rangle|^2 = \left(-\tfrac{3}{\sqrt6}\right)^2 = \tfrac32$ .

Problem 4 · Inner Products from Coordinates

Given: $e_1, e_2$ is an orthonormal basis of a complex inner product space $V$ , and $u = 2e_1 + i\,e_2$ , $\ v = (1 + i)e_1 + 3e_2$ . Find $\langle u, v \rangle$ .

✅ Correct!

\langle u, v \rangle = 2 \cdot \overline{1 + i} + i \cdot \overline{3} = 2 - 2i + 3i = 2 + i

❌ Close, but…

2(1 + i) + 3i

forgets the conjugate: a scalar pulled out of the second slot comes out conjugated, so the formula uses

\overline{\langle v, e_k \rangle}

❌ Close, but… That conjugates the coordinates of

u

; the conjugates belong to the coordinates of

v

, the vector in the second slot.

❌ Not quite. Only the coordinates of

v

are conjugated, not those of

u

as well.

❌ Not quite. Use

\langle u, v \rangle = \langle u, e_1 \rangle \overline{\langle v, e_1 \rangle} + \langle u, e_2 \rangle \overline{\langle v, e_2 \rangle}

Show solution

In an orthonormal basis the coefficients are the inner products, so $\langle u, e_1 \rangle = 2$ , $\langle u, e_2 \rangle = i$ , $\langle v, e_1 \rangle = 1 + i$ , $\langle v, e_2 \rangle = 3$ . The third part of the theorem gives

\langle u, v \rangle = \langle u, e_1 \rangle \overline{\langle v, e_1 \rangle} + \langle u, e_2 \rangle \overline{\langle v, e_2 \rangle} = 2(1 - i) + i \cdot 3 = 2 - 2i + 3i = 2 + i.

It comes from taking the inner product of $u$ with $v = \langle v, e_1 \rangle e_1 + \langle v, e_2 \rangle e_2$ : the second slot is additive, and each scalar $\langle v, e_k \rangle$ comes out of it conjugated.

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