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9.2Estimated reading time: 8 min

9.2 Orthogonal sets and orthonormal bases

Use orthogonality to build orthogonal and orthonormal bases, then read coefficients without solving a linear system every time.

Course contents

MATH1030: Linear algebra I

Rigorous linear algebra notes on systems, matrices, structure, and proof, with interaction used only where it clarifies the mathematics.

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All note collectionsSeries overview
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Chapter 1Systems of equations1 sections
1.11.1 Equations and solution sets
Chapter 2Matrices and elimination4 sections
2.12.1 Matrix basics2.22.2 Augmented matrices and row operations2.32.3 Gaussian elimination and RREF2.42.4 Solution-set types
Chapter 3Matrix algebra2 sections
3.13.1 Matrix multiplication and identity matrices3.23.2 Transpose and special matrices
Chapter 4Solution structure1 sections
4.14.1 Homogeneous systems and null space
Chapter 5Invertibility1 sections
5.15.1 Invertible matrices
Chapter 6Vector spaces6 sections
6.16.1 Vector spaces6.26.2 Subspaces6.36.3 Linear combinations and span6.46.4 Linear dependence and independence6.56.5 Basis and dimension6.66.6 Column space, row space, and rank
Chapter 7Determinants3 sections
7.17.1 Determinants and cofactor expansion7.27.2 Row operations, products, and invertibility7.37.3 Transpose, column operations, and Cramer's rule
Chapter 8Eigenvalues and diagonalization3 sections
8.18.1 Eigenvalues, eigenvectors, and eigenspaces8.28.2 Diagonalization and similarity8.38.3 Characteristic polynomials and diagonalization tests
Chapter 9Inner products and orthogonality4 sections
9.19.1 Inner products, norms, and angles9.29.2 Orthogonal sets and orthonormal basesCurrent section9.39.3 Gram-Schmidt orthogonalization9.49.4 Cauchy-Schwarz and triangle inequalities

Once an inner product is available, you can single out vector families whose geometry is especially simple. Orthogonal sets are the first major example.

They matter because orthogonality replaces elimination. In an orthogonal basis, coordinates can be read off directly from inner products instead of solving a new linear system every time.

Orthogonality

Definition

Orthogonal vectors

Two vectors v,w∈Rmv,w\in\mathbb{R}^m are orthogonal if

⟨v,w⟩=0.\langle v,w\rangle=0.

This is written as

v⊥w.v\perp w.

The point is not merely that the vectors look perpendicular in a picture. The definition works in every dimension because it is stated algebraically through the inner product.

Worked example

Check orthogonality by inner product

Let

v=[123],w=[−1−11].v= \begin{bmatrix} 1\\2\\3 \end{bmatrix}, \qquad w= \begin{bmatrix} -1\\-1\\1 \end{bmatrix}.

Then

⟨v,w⟩=1(−1)+2(−1)+3(1)=0.\langle v,w\rangle=1(-1)+2(-1)+3(1)=0.

So v and w are orthogonal.

Orthogonal sets

Definition

Orthogonal set

A finite set S={v1,…,vk}S=\{v_1,\dots,v_k\} in Rm\mathbb{R}^m is orthogonal if:

  1. every vector in the set is nonzero;
  2. vi⊥vjv_i\perp v_j whenever i≠ji\neq j.

The nonzero requirement matters. If zero were allowed, every set containing 0 would be vacuously orthogonal, which would destroy the structural theorems.

Theorem

Orthogonal nonzero sets are automatically linearly independent

If S={v1,…,vk}S=\{v_1,\dots,v_k\} is an orthogonal subset of Rm\mathbb{R}^m, then SS is linearly independent.

Proof

Why orthogonal sets are independent

Coordinates in an orthogonal basis

Orthogonality is powerful because it makes coordinates explicit.

Theorem

Coefficient formula in an orthogonal set

Suppose S={v1,…,vk}S=\{v_1,\dots,v_k\} is an orthogonal set and

v=α1v1+⋯+αkvk.v=\alpha_1v_1+\cdots+\alpha_kv_k.

Then for each i,

αi=⟨v,vi⟩∥vi∥2.\alpha_i=\frac{\langle v,v_i\rangle}{\|v_i\|^2}.

Hence

v=⟨v,v1⟩∥v1∥2v1+⋯+⟨v,vk⟩∥vk∥2vk.v= \frac{\langle v,v_1\rangle}{\|v_1\|^2}v_1 +\cdots+ \frac{\langle v,v_k\rangle}{\|v_k\|^2}v_k.

This formula is one of the main practical rewards of orthogonality. There is no need to solve a system to recover the coefficients.

Worked example

Recover coordinates from inner products

Let

v1=[111],v2=[1−10],v3=[11−2].v_1= \begin{bmatrix} 1\\1\\1 \end{bmatrix}, \quad v_2= \begin{bmatrix} 1\\-1\\0 \end{bmatrix}, \quad v_3= \begin{bmatrix} 1\\1\\-2 \end{bmatrix}.

This is an orthogonal set. Let

v=[123].v= \begin{bmatrix} 1\\2\\3 \end{bmatrix}.

Then

α1=⟨v,v1⟩∥v1∥2=63=2,\alpha_1=\frac{\langle v,v_1\rangle}{\|v_1\|^2} =\frac{6}{3}=2,α2=⟨v,v2⟩∥v2∥2=−12=−12,\alpha_2=\frac{\langle v,v_2\rangle}{\|v_2\|^2} =\frac{-1}{2}=-\frac12,α3=⟨v,v3⟩∥v3∥2=−36=−12.\alpha_3=\frac{\langle v,v_3\rangle}{\|v_3\|^2} =\frac{-3}{6}=-\frac12.

So

v=2v1−12v2−12v3.v=2v_1-\frac12v_2-\frac12v_3.

Orthogonal and orthonormal bases

Definition

Orthogonal basis

Let VV be a subspace of Rm\mathbb{R}^m. A subset SS of VV is an orthogonal basis for VV if SS is both:

  1. a basis of VV;
  2. an orthogonal set.

Because orthogonal nonzero sets are automatically linearly independent, to check that an orthogonal set is an orthogonal basis you only need to check that it spans the subspace.

Theorem

Orthogonal basis criterion

Let VV be a subspace of Rm\mathbb{R}^m. If SS is an orthogonal subset of VV, then SS is an orthogonal basis for VV if and only if SS spans VV.

Definition

Orthonormal basis

A set S={v1,…,vk}S=\{v_1,\dots,v_k\} is orthonormal if it is orthogonal and every vector has norm 1. Equivalently,

⟨vi,vj⟩={1,i=j,0,i≠j.\langle v_i,v_j\rangle= \begin{cases} 1,& i=j,\\ 0,& i\neq j. \end{cases}

An orthonormal set that is also a basis is an orthonormal basis.

For orthonormal bases, the coordinate formula becomes even cleaner.

Theorem

Coordinate formula in an orthonormal basis

If S={v1,…,vk}S=\{v_1,\dots,v_k\} is an orthonormal set and

v∈span⁡{v1,…,vk},v\in\operatorname{span}\{v_1,\dots,v_k\},

then

v=⟨v,v1⟩v1+⋯+⟨v,vk⟩vk.v=\langle v,v_1\rangle v_1+\cdots+\langle v,v_k\rangle v_k.

The denominators disappear because every norm is 1.

Common mistake

Common mistake

Orthogonal does not mean 'already a basis'

An orthogonal set is automatically linearly independent, but it may still fail to span the space you care about. Orthogonality gives independence for free; it does not give spanning for free.

Quick check

Quick check

Can an orthogonal set contain the zero vector?

Use the definition, not only the equation ⟨0,v⟩=0⟨0,v⟩=0.

Solution

Answer

Quick check

Why are the standard basis vectors e_1,\dots,e_m orthogonal?

Compute their pairwise inner products.

Solution

Answer

Quick check

If an orthonormal basis is used, what is the coefficient of v_i in the expansion of v?

Use the orthonormal coordinate formula.

Solution

Answer

Exercises

Quick check

Show that {(1,2),(−2,1)}\{(1,2),(-2,1)\} is an orthogonal basis of R2\mathbb{R}^2.

Check orthogonality first, then spanning.

Solution

Guided solution

Quick check

Let u1=(1,0,1)u_1=(1,0,1) and u2=(1,−2,1)u_2=(1,-2,1). Find the coefficients of v=(2,−2,2)v=(2,-2,2) in this orthogonal basis.

Use the orthogonal coefficient formula directly.

Solution

Guided solution

Quick check

Normalize the orthogonal basis {(3,4),(4,−3)}\{(3,4),(4,-3)\}.

Divide each vector by its norm.

Solution

Guided solution

Related notes

Read 9.1 Inner products, norms, and angles first if the inner-product formulas themselves are not yet automatic.

Continue with 9.3 Gram-Schmidt orthogonalization to see how arbitrary bases are converted into orthogonal ones.

The basis language here also depends on 6.5 Basis and dimension.

Section mastery checkpoint

Answer each question correctly to complete this section checkpoint. Correct progress: 0%.

Skills: orthonormal-basis, inner-product, coordinates

Fill in the blank: if {v1,...,vk} is orthonormal and v lies in its span, then the coefficient of vi in the expansion of v is ____.

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Prerequisites

9.1

9.1 Inner products, norms, and angles

6.5

6.5 Basis and dimension

Key terms in this unit

More notes in this series

Previous section

9.1 Inner products, norms, and angles

Next section

9.3 Gram-Schmidt orthogonalization