Orthonormal Basis

linear-algebra

Definition

Orthonormal Basis

Let $(V,⟨\cdot |\cdot ⟩)$ be an inner product space. A family of vectors $\{e_i{\}}_{i\in I}\subseteq V$ is an orthonormal basis if:

1. Orthogonality: $⟨e_i|e_j⟩=0$ for all $i\ne j$
2. Normalisation: $⟨e_i|e_i⟩=1$ for all $i\in I$
3. Completeness: $\overline{\mathrm{span}\{e_i:i\in I\}}=V$

where $⟨\cdot |\cdot ⟩$ is the bra-ket notation and $\text{span}(\cdot )$ denotes the linear span.

Every $v\in V$ then admits the expansion

$$v=\sum _i⟨e_i|v⟩e_i$$

Kronecker Delta

The orthonormality condition can be written as

$$⟨e_i|e_j⟩={\delta }_{ij}$$

for all $i,j\in I$, where ${\delta }_{ij}$ is the Kronecker delta. This means the basis vectors are mutually orthogonal and each vector has norm $1$.

Parseval Identity

If $c_i=⟨e_i|v⟩$, then the expansion coefficients satisfy

$$\Vert v{\Vert }^2=\sum _i|c_i{|}^2$$

This is Parseval’s identity. It states that the squared norm of a vector equals the sum of the squared magnitudes of its coordinates in an orthonormal basis.

Gram-Schmidt

Gram-Schmidt

Given a linearly independent basis, we can form linear combinations of the basis vectors to obtain an orthonormal basis.

Example

Trivial Orthonormal Basis of $\mathbb{R}$

The trivial orthonormal basis of $\mathbb{R}$ is the family $\{1\}$. With the usual inner product on $\mathbb{R}$,

$$⟨1,1⟩=1.$$

Every $x\in \mathbb{R}$ can then be written as

$$x=⟨1,x⟩\cdot 1=x\cdot 1=x.$$

Understanding

1. The basis has only one vector, $e_1=1$.
2. The coefficient of $x$ in this basis is $c_1=⟨e_1,x⟩=⟨1,x⟩$.
3. In $\mathbb{R}$ with the usual inner product, $⟨1,x⟩=x$.
4. Substituting into the expansion formula gives

$$x=c_1{e}_1=⟨1,x⟩\cdot 1.$$

Orthonormal Basis of ${\mathbb{R}}^2$

The standard basis of ${\mathbb{R}}^2$ is the family

$$\{e_1,e_2\}=\left\{\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)\right\}.$$

With the usual inner product,

$$⟨e_1,e_1⟩=1,⟨e_2,e_2⟩=1,⟨e_1,e_2⟩=0.$$

So every vector

$$v=\left(\begin{array}{c}x\\ y\end{array}\right)$$

can be written as

$$v=⟨e_1,v⟩e_1+⟨e_2,v⟩e_2=x{e}_1+y{e}_2.$$

Understanding

1. The basis vectors are the unit vectors along the coordinate axes.
2. The coefficient of $v$ along $e_1$ is $⟨e_1,v⟩=x$.
3. The coefficient of $v$ along $e_2$ is $⟨e_2,v⟩=y$.
4. Substituting these coefficients gives the usual coordinate decomposition of $v$.