Orthonormal Basis in Linear Algebra

An orthonormal basis combines two vital properties: every vector in the set is a unit vector (having a length or norm equal to 1), and all distinct vectors are mutually orthogonal to each other. Sets of orthogonal vectors are inherently linearly independent.

Table of Contents

Orthonormal Basis Definition

A set of vectors \( \{ \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \} \) in an inner product space is called an orthonormal set if:

  1. Unit Vectors (Normalization): \( \|\textbf{u}_i\| = 1 \) for all \( i \)
  2. Orthogonality: \( \textbf{u}_i \cdot \textbf{u}_j = 0 \) whenever \( i \neq j \)

If this set also spans the vector space (forming a basis), it is called an orthonormal basis.

Properties of Orthonormal Bases

Orthonormal bases have several important properties that make them especially useful in linear algebra:

  1. Linear independence: Any orthonormal set is automatically linearly independent.
  2. Coordinates via dot product: If \( \{ \textbf{u}_1, \dots, \textbf{u}_n \} \) is an orthonormal basis, then any vector \( \textbf{v} \) can be written as \( \textbf{v} = (\textbf{v} \cdot \textbf{u}_1)\textbf{u}_1 + \dots + (\textbf{v} \cdot \textbf{u}_n)\textbf{u}_n \).
  3. Preservation of length: For an orthonormal basis, \( \|\textbf{v}\|^2 = \sum_{i=1}^{n} (\textbf{v} \cdot \textbf{u}_i)^2 \).
  4. Unique representation: Every vector in the space has a unique representation as a linear combination of the basis vectors.
  5. Simplified projections: The projection of a vector onto an orthonormal basis vector simplifies to \( \text{proj}_{\textbf{u}_i}(\textbf{v}) = (\textbf{v} \cdot \textbf{u}_i)\textbf{u}_i \).

Note: The Gram-Schmidt process can transform any basis into an orthonormal basis.

Examples with Solutions

Example 1: The Standard Basis in \( \mathbb{R}^3 \)

Show Solution

Show that the standard unit vectors \( \textbf{i} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( \textbf{j} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), and \( \textbf{k} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) form an orthonormal basis.

Solution

First, show that the inner product of each pair of vectors is equal to \( 0 \):

\[ \textbf{i} \cdot \textbf{j} = (1)(0) + (0)(1) + (0)(0) = 0 \] \[ \textbf{i} \cdot \textbf{k} = (1)(0) + (0)(0) + (0)(1) = 0 \] \[ \textbf{j} \cdot \textbf{k} = (0)(0) + (1)(0) + (0)(1) = 0 \]

Thus, the three vectors are mutually orthogonal (and consequently linearly independent).

Next, show that the length (norm) of each vector is equal to \( 1 \):

\[ \|\textbf{i}\| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1 \] \[ \|\textbf{j}\| = \sqrt{0^2 + 1^2 + 0^2} = \sqrt{1} = 1 \] \[ \|\textbf{k}\| = \sqrt{0^2 + 0^2 + 1^2} = \sqrt{1} = 1 \]

Since the vectors are mutually orthogonal and each has a length of \( 1 \), they form an orthonormal basis for \( \mathbb{R}^3 \).

Example 2: Scaling Vectors to Form an Orthonormal Basis

Show Solution

a) Show that the vectors \( \textbf{u} = K \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \), \( \textbf{v} = L \begin{bmatrix} 4 \\ 1 \\ -2 \end{bmatrix} \), and \( \textbf{w} = M \begin{bmatrix} -2 \\ 10 \\ 1 \end{bmatrix} \), where \( K, L, M \) are real constants, are orthogonal.
b) Find positive values for the constants \( K, L, \) and \( M \) so that \( \textbf{u}, \textbf{v}, \) and \( \textbf{w} \) form an orthonormal basis.

Solution

a) Calculate the inner product of each pair of vectors:

\[ \textbf{u} \cdot \textbf{v} = KL[(1)(4) + (0)(1) + (2)(-2)] = KL(4 + 0 - 4) = 0 \] \[ \textbf{u} \cdot \textbf{w} = KM[(1)(-2) + (0)(10) + (2)(1)] = KM(-2 + 0 + 2) = 0 \] \[ \textbf{v} \cdot \textbf{w} = LM[(4)(-2) + (1)(10) + (-2)(1)] = LM(-8 + 10 - 2) = 0 \]

Therefore, vectors \( \textbf{u}, \textbf{v}, \) and \( \textbf{w} \) are orthogonal for any real values of \( K, L, M \).

b) To make them an orthonormal basis, we enforce that the norm of each vector equals \( 1 \):

For \( \textbf{u} \):

\[ \|\textbf{u}\| = \sqrt{(K)^2 + 0^2 + (2K)^2} = \sqrt{K^2 + 4K^2} = \sqrt{5K^2} = |K|\sqrt{5} = 1 \]

Since \( K \) is positive, \( K\sqrt{5} = 1 \implies K = \frac{1}{\sqrt{5}} \).

For \( \textbf{v} \):

\[ \|\textbf{v}\| = \sqrt{(4L)^2 + L^2 + (-2L)^2} = \sqrt{16L^2 + L^2 + 4L^2} = \sqrt{21L^2} = |L|\sqrt{21} = 1 \]

Solving for positive \( L \): \( L = \frac{1}{\sqrt{21}} \).

For \( \textbf{w} \):

\[ \|\textbf{w}\| = \sqrt{(-2M)^2 + (10M)^2 + (M)^2} = \sqrt{4M^2 + 100M^2 + M^2} = \sqrt{105M^2} = |M|\sqrt{105} = 1 \]

Solving for positive \( M \): \( M = \frac{1}{\sqrt{105}} \).

With these values, \( \textbf{u}, \textbf{v}, \) and \( \textbf{w} \) form an orthonormal basis for \( \mathbb{R}^3 \).

Example 3: Proving Trigonometric Vectors Form an Orthonormal Basis

Show Solution

Prove that the vectors \( \textbf{u} = \begin{bmatrix} \sin(\theta)\cos(\phi) \\ \sin(\theta)\sin(\phi) \\ \cos(\theta) \end{bmatrix} \), \( \textbf{v} = \begin{bmatrix} \cos(\theta)\cos(\phi) \\ \cos(\theta)\sin(\phi) \\ -\sin(\theta) \end{bmatrix} \), and \( \textbf{w} = \begin{bmatrix} -\sin(\phi) \\ \cos(\phi) \\ 0 \end{bmatrix} \) form an orthonormal basis.

Solution

In Example 6 of orthogonal vectors, it was already proved that \( \textbf{u}, \textbf{v}, \) and \( \textbf{w} \) are mutually orthogonal. We now only need to show that the norm of each vector is equal to \( 1 \).

1) Norm of \( \textbf{u} \):

\[ \|\textbf{u}\| = \sqrt{(\sin(\theta)\cos(\phi))^2 + (\sin(\theta)\sin(\phi))^2 + (\cos(\theta))^2} \] \[ \|\textbf{u}\| = \sqrt{\sin^2(\theta)\cos^2(\phi) + \sin^2(\theta)\sin^2(\phi) + \cos^2(\theta)} \]

Factor out \( \sin^2(\theta) \) from the first two terms:

\[ \|\textbf{u}\| = \sqrt{\sin^2(\theta)[\cos^2(\phi) + \sin^2(\phi)] + \cos^2(\theta)} \]

Using the identity \( \cos^2(\phi) + \sin^2(\phi) = 1 \):

\[ \|\textbf{u}\| = \sqrt{\sin^2(\theta)(1) + \cos^2(\theta)} = \sqrt{\sin^2(\theta) + \cos^2(\theta)} = \sqrt{1} = 1 \]

2) Norm of \( \textbf{v} \):

\[ \|\textbf{v}\| = \sqrt{(\cos(\theta)\cos(\phi))^2 + (\cos(\theta)\sin(\phi))^2 + (-\sin(\theta))^2} \] \[ \|\textbf{v}\| = \sqrt{\cos^2(\theta)\cos^2(\phi) + \cos^2(\theta)\sin^2(\phi) + \sin^2(\theta)} \]

Factor out \( \cos^2(\theta) \) from the first two terms and apply the trigonometric identity:

\[ \|\textbf{v}\| = \sqrt{\cos^2(\theta)[\cos^2(\phi) + \sin^2(\phi)] + \sin^2(\theta)} = \sqrt{\cos^2(\theta) + \sin^2(\theta)} = 1 \]

3) Norm of \( \textbf{w} \):

\[ \|\textbf{w}\| = \sqrt{(-\sin(\phi))^2 + (\cos(\phi))^2 + 0^2} = \sqrt{\sin^2(\phi) + \cos^2(\phi)} = 1 \]

Since the vectors are mutually orthogonal and each has a norm of \( 1 \), they form an orthonormal basis. Geometrically, these vectors correspond to the standard orthonormal basis in spherical coordinates.

Frequently Asked Questions

What is an orthonormal basis?

An orthonormal basis is a set of vectors that are all mutually orthogonal (perpendicular), each have a length (norm) of 1, and span the entire vector space.

What is the difference between orthogonal and orthonormal?

Orthogonal vectors are perpendicular to each other (dot product equals zero). Orthonormal vectors are orthogonal AND each has a length of 1.

How do you normalize a vector?

To normalize a vector, divide it by its norm (length): \( \hat{\textbf{u}} = \frac{\textbf{u}}{\|\textbf{u}\|} \). The resulting vector has a length of 1 and points in the same direction.

Why are orthonormal bases useful?

Orthonormal bases simplify many calculations: finding coordinates only requires dot products, projections are easier to compute, and numerical stability is improved.

How do you convert any basis into an orthonormal basis?

The Gram-Schmidt process takes any basis and produces an orthogonal basis, which can then be normalized to obtain an orthonormal basis.

More References and Links