Orthonormality

What does a pair of orthonormal vectors in 2-D Euclidean space look like?

Let u = (x₁, y₁) and v = (x₂, y₂). Consider the restrictions on x₁, x₂, y₁, y₂ required to make u and v form an orthonormal pair.

• From the orthogonality restriction, u • v = 0.
• From the unit length restriction on u, ||u|| = 1.
• From the unit length restriction on v, ||v|| = 1.

Expanding these terms gives 3 equations:

1. ${\displaystyle x_{1}x_{2}+y_{1}y_{2}=0\quad }$
2. ${\displaystyle {\sqrt {{x_{1}}^{2}+{y_{1}}^{2}}}=1}$
3. ${\displaystyle {\sqrt {{x_{2}}^{2}+{y_{2}}^{2}}}=1}$

Converting from Cartesian to polar coordinates, and considering Equation ${\displaystyle (2)}$ and Equation ${\displaystyle (3)}$ immediately gives the result r₁ = r₂ = 1. In other words, requiring the vectors be of unit length restricts the vectors to lie on the unit circle.

After substitution, Equation ${\displaystyle (1)}$ becomes ${\displaystyle \cos \theta _{1}\cos \theta _{2}+\sin \theta _{1}\sin \theta _{2}=0}$. Rearranging gives ${\displaystyle \tan \theta _{1}=-\cot \theta _{2}}$. Using a trigonometric identity to convert the cotangent term gives

${\displaystyle \tan(\theta _{1})=\tan \left(\theta _{2}+{\tfrac {\pi }{2}}\right)}$

${\displaystyle \Rightarrow \theta _{1}=\theta _{2}+{\tfrac {\pi }{2}}}$

It is clear that in the plane, orthonormal vectors are simply radii of the unit circle whose difference in angles equals 90°.

Orthonormal sets have certain very appealing properties, which make them particularly easy to work with.

• Theorem. If {e₁, e₂, ..., e_n} is an orthonormal list of vectors, then
  $${\displaystyle \forall {\textbf {a}}:=[a_{1},\cdots ,a_{n}];\ \|a_{1}{\textbf {e}}_{1}+a_{2}{\textbf {e}}_{2}+\cdots +a_{n}{\textbf {e}}_{n}\|^{2}=|a_{1}|^{2}+|a_{2}|^{2}+\cdots +|a_{n}|^{2}}$$

  
• Theorem. Every orthonormal list of vectors is linearly independent.

• Gram-Schmidt theorem. If {v₁, v₂,...,v_n} is a linearly independent list of vectors in an inner-product space ${\displaystyle {\mathcal {V}}}$, then there exists an orthonormal list {e₁, e₂,...,e_n} of vectors in ${\displaystyle {\mathcal {V}}}$ such that span(e₁, e₂,...,e_n) = span(v₁, v₂,...,v_n).

Proof of the Gram-Schmidt theorem is constructive, and discussed at length elsewhere. The Gram-Schmidt theorem, together with the axiom of choice, guarantees that every vector space admits an orthonormal basis. This is possibly the most significant use of orthonormality, as this fact permits operators on inner-product spaces to be discussed in terms of their action on the space's orthonormal basis vectors. What results is a deep relationship between the diagonalizability of an operator and how it acts on the orthonormal basis vectors. This relationship is characterized by the Spectral Theorem.

The standard basis for the coordinate space Fⁿ is

{e1, e2,...,en}   where	e1 = (1, 0, ..., 0)
	e2 = (0, 1, ..., 0)
	${\displaystyle \vdots }$
	en = (0, 0, ..., 1)

Any two vectors e_i, e_j where i≠j are orthogonal, and all vectors are clearly of unit length. So {e₁, e₂,...,e_n} forms an orthonormal basis.

When referring to real-valued functions, usually the L² inner product is assumed unless otherwise stated. Two functions ${\displaystyle \phi (x)}$ and ${\displaystyle \psi (x)}$ are orthonormal over the interval ${\displaystyle [a,b]}$ if

${\displaystyle (1)\quad \langle \phi (x),\psi (x)\rangle =\int _{a}^{b}\phi (x)\psi (x)dx=0,\quad {\rm {and}}}$

${\displaystyle (2)\quad ||\phi (x)||_{2}=||\psi (x)||_{2}=\left[\int _{a}^{b}|\phi (x)|^{2}dx\right]^{\frac {1}{2}}=\left[\int _{a}^{b}|\psi (x)|^{2}dx\right]^{\frac {1}{2}}=1.}$

The Fourier series is a method of expressing a periodic function in terms of sinusoidal basis functions. Taking C[−π,π] to be the space of all real-valued functions continuous on the interval [−π,π] and taking the inner product to be

${\displaystyle \langle f,g\rangle =\int _{-\pi }^{\pi }f(x)g(x)dx}$

it can be shown that

${\displaystyle \left\{{\frac {1}{\sqrt {2\pi }}},{\frac {\sin(x)}{\sqrt {\pi }}},{\frac {\sin(2x)}{\sqrt {\pi }}},\ldots ,{\frac {\sin(nx)}{\sqrt {\pi }}},{\frac {\cos(x)}{\sqrt {\pi }}},{\frac {\cos(2x)}{\sqrt {\pi }}},\ldots ,{\frac {\cos(nx)}{\sqrt {\pi }}}\right\},\quad n\in \mathbb {N} }$

forms an orthonormal set.

However, this is of little consequence, because C[−π,π] is infinite-dimensional, and a finite set of vectors cannot span it. But, removing the restriction that n be finite makes the set dense in C[−π,π] and therefore an orthonormal basis of C[−π,π].