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Orthonormal Basis — Definition, Formula & Examples

An orthonormal basis is a set of vectors that are all mutually perpendicular (orthogonal) and each have a magnitude of 1. These vectors span the entire space, so any vector in that space can be written as a unique combination of them.

A set of vectors {e1,e2,…,en}\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\} forms an orthonormal basis for an inner product space VV of dimension nn if the vectors are pairwise orthogonal (⟨ei,ej⟩=0\langle \mathbf{e}_i, \mathbf{e}_j \rangle = 0 for i≠ji \neq j), each vector is unit length (⟨ei,ei⟩=1\langle \mathbf{e}_i, \mathbf{e}_i \rangle = 1), and the set spans VV.

Key Formula

⟨ei,ej⟩=δij={1if i=j0if i≠j\langle \mathbf{e}_i, \mathbf{e}_j \rangle = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}
Where:
  • ei,ej\mathbf{e}_i, \mathbf{e}_j = Vectors in the orthonormal basis
  • ⟨⋅,⋅⟩\langle \cdot, \cdot \rangle = Inner product (e.g., the dot product in Euclidean space)
  • δij\delta_{ij} = Kronecker delta, equals 1 when indices match and 0 otherwise

How It Works

To check whether a set of vectors is an orthonormal basis, verify two things: every pair of distinct vectors has a dot product of zero, and each vector has magnitude 1. The standard basis in R3\mathbb{R}^3, namely {i^,j^,k^}\{\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\}, is the most familiar example. When you have an orthonormal basis, finding a vector's coordinates becomes simple: the coefficient for each basis vector is just the dot product of your vector with that basis vector. This eliminates the need to solve systems of equations. The Gram-Schmidt process is the standard algorithm for converting any basis into an orthonormal one.

Worked Example

Problem: Verify that the set {e1=(12,12),  e2=(−12,12)}\left\{\mathbf{e}_1 = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),\; \mathbf{e}_2 = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\right\} is an orthonormal basis for R2\mathbb{R}^2, then express v=(3,1)\mathbf{v} = (3, 1) in this basis.
Check orthogonality: Compute the dot product of the two vectors.
e1⋅e2=12(−12)+12(12)=−12+12=0\mathbf{e}_1 \cdot \mathbf{e}_2 = \frac{1}{\sqrt{2}}\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2} + \frac{1}{2} = 0
Check unit length: Compute the magnitude of each vector.
∥e1∥=12+12=1,∥e2∥=12+12=1\|\mathbf{e}_1\| = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1, \quad \|\mathbf{e}_2\| = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1
Find coordinates of v: Since the basis is orthonormal, each coordinate is just the dot product of v with the corresponding basis vector.
c_1 = \mathbf{v} \cdot \mathbf{e}_1 = \frac{3}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$$$c_2 = \mathbf{v} \cdot \mathbf{e}_2 = -\frac{3}{\sqrt{2}} + \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}
Answer: The set is orthonormal, and v=22 e1−2 e2\mathbf{v} = 2\sqrt{2}\,\mathbf{e}_1 - \sqrt{2}\,\mathbf{e}_2.

Why It Matters

Orthonormal bases are essential in quantum mechanics, signal processing (Fourier series are built on orthonormal functions), and computer graphics. They make projections, least-squares approximations, and coordinate transformations computationally simple and numerically stable.

Common Mistakes

Mistake: Confusing orthogonal with orthonormal. A set of mutually perpendicular vectors that do not all have magnitude 1 is orthogonal but not orthonormal.
Correction: Always verify both conditions: pairwise orthogonality and unit length. If the vectors are orthogonal but not unit length, normalize each one by dividing it by its magnitude.