Orthonormal basis

• For ${\displaystyle \mathbb {R} ^{3}}$, the set of vectors ${\displaystyle \left\{e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)\right\},}$ is called the standard basis and forms an orthonormal basis of ${\displaystyle \mathbb {R} ^{3}}$ with respect to the standard dot product. Note that both the standard basis and standard dot product rely on viewing ${\displaystyle \mathbb {R} ^{3}}$ as the Cartesian product ${\displaystyle \mathbb {R} \times \mathbb {R} \times \mathbb {R} }$

  Proof: A straightforward computation shows that the inner products of these vectors equals zero, ${\displaystyle \left\langle e_{1},e_{2}\right\rangle =\left\langle e_{1},e_{3}\right\rangle =\left\langle e_{2},e_{3}\right\rangle =0}$ and that each of their magnitudes equals one, ${\displaystyle \left\|e_{1}\right\|=\left\|e_{2}\right\|=\left\|e_{3}\right\|=1.}$ This means that ${\displaystyle \left\{e_{1},e_{2},e_{3}\right\}}$ is an orthonormal set. All vectors ${\displaystyle (x,y,z)\in \mathbb {R} ^{3}}$ can be expressed as a sum of the basis vectors scaled

  $${\displaystyle (x,y,z)=xe_{1}+ye_{2}+ze_{3},}$$

  

  so ${\displaystyle \left\{e_{1},e_{2},e_{3}\right\}}$ spans ${\displaystyle \mathbb {R} ^{3}}$ and hence must be a basis. It may also be shown that the standard basis rotated about an axis through the origin or reflected in a plane through the origin also forms an orthonormal basis of ${\displaystyle \mathbb {R} ^{3}}$.

• For ${\displaystyle \mathbb {R} ^{n}}$, the standard basis and inner product are similarly defined. Any other orthonormal basis is related to the standard basis by an orthogonal transformation in the group O(n).
• For pseudo-Euclidean space ${\displaystyle \mathbb {R} ^{p,q},}$, an orthogonal basis ${\displaystyle \{e_{\mu }\}}$ with metric ${\displaystyle \eta }$ instead satisfies ${\displaystyle \eta (e_{\mu },e_{\nu })=0}$ if ${\displaystyle \mu \neq \nu }$, ${\displaystyle \eta (e_{\mu },e_{\mu })=+1}$ if ${\displaystyle 1\leq \mu \leq p}$, and ${\displaystyle \eta (e_{\mu },e_{\mu })=-1}$ if ${\displaystyle p+1\leq \mu \leq p+q}$. Any two orthonormal bases are related by a pseudo-orthogonal transformation. In the case ${\displaystyle (p,q)=(1,3)}$, these are Lorentz transformations.
• The set ${\displaystyle \left\{f_{n}:n\in \mathbb {Z} \right\}}$ with ${\displaystyle f_{n}(x)=\exp(2\pi inx),}$ where ${\displaystyle \exp }$ denotes the exponential function, forms an orthonormal basis of the space of functions with finite Lebesgue integrals, ${\displaystyle L^{2}([0,1]),}$ with respect to the 2-norm. This is fundamental to the study of Fourier series.
• The set ${\displaystyle \left\{e_{b}:b\in B\right\}}$ with ${\displaystyle e_{b}(c)=1}$ if ${\displaystyle b=c}$ and ${\displaystyle e_{b}(c)=0}$ otherwise forms an orthonormal basis of ${\displaystyle \ell ^{2}(B).}$
• Eigenfunctions of a Sturm–Liouville eigenproblem.
• The column vectors of an orthogonal matrix form an orthonormal set.

If ${\displaystyle B}$ is an orthogonal basis of ${\displaystyle H,}$ then every element ${\displaystyle x\in H}$ may be written as

$${\displaystyle x=\sum _{b\in B}{\frac {\langle b,x\rangle }{\lVert b\rVert ^{2}}}b.}$$

When ${\displaystyle B}$ is orthonormal, this simplifies to

$${\displaystyle x=\sum _{b\in B}\langle b,x\rangle b}$$

and the square of the norm of ${\displaystyle x}$ can be given by

$${\displaystyle \|x\|^{2}=\sum _{b\in B}|\langle x,b\rangle |^{2}.}$$

Even if ${\displaystyle B}$ is uncountable, only countably many terms in this sum will be non-zero, and the expression is therefore well-defined. This sum is also called the Fourier expansion of ${\displaystyle x,}$ and the formula is usually known as Parseval's identity.

If ${\displaystyle B}$ is an orthonormal basis of ${\displaystyle H,}$ then ${\displaystyle H}$ is isomorphic to ${\displaystyle \ell ^{2}(B)}$ in the following sense: there exists a bijective linear map ${\displaystyle \Phi :H\to \ell ^{2}(B)}$such that

$${\displaystyle \langle \Phi (x),\Phi (y)\rangle =\langle x,y\rangle \quad {\text{ for all }}x,y\in H.}$$

Given a Hilbert space ${\displaystyle H}$ and a set ${\displaystyle S}$ of mutually orthogonal vectors in ${\displaystyle H,}$ we can take the smallest closed linear subspace ${\displaystyle V}$ of ${\displaystyle H}$ containing ${\displaystyle S.}$ Then ${\displaystyle S}$ will be an orthogonal basis of ${\displaystyle V;}$ which may of course be smaller than ${\displaystyle H}$ itself, being an incomplete orthogonal set, or be ${\displaystyle H,}$ when it is a complete orthogonal set.

Using Zorn's lemma and the Gram–Schmidt process (or more simply well-ordering and transfinite recursion), one can show that every Hilbert space admits an orthonormal basis;[6] furthermore, any two orthonormal bases of the same space have the same cardinality (this can be proven in a manner akin to that of the proof of the usual dimension theorem for vector spaces, with separate cases depending on whether the larger basis candidate is countable or not). A Hilbert space is separable if and only if it admits a countable orthonormal basis. (One can prove this last statement without using the axiom of choice.)

For concreteness we discuss orthonormal bases for a real, ${\displaystyle n}$ dimensional vector space ${\displaystyle V}$ with a positive definite symmetric bilinear form ${\displaystyle \phi =\langle \cdot ,\cdot \rangle }$.

One way to view an orthonormal basis with respect to ${\displaystyle \phi }$ is as a set of vectors ${\displaystyle {\mathcal {B}}=\{e_{i}\}}$, which allow us to write ${\displaystyle v=v^{i}e_{i}}$ for ${\displaystyle v\in V}$, and ${\displaystyle v^{i}\in \mathbb {R} }$ or ${\displaystyle (v^{i})\in \mathbb {R} ^{n}}$. With respect to this basis, the components of ${\displaystyle \phi }$ are particularly simple: ${\displaystyle \phi (e_{i},e_{j})=\delta _{ij}.}$

We can now view the basis as a map ${\displaystyle \psi _{\mathcal {B}}:V\rightarrow \mathbb {R} ^{n}}$ which is an isomorphism of inner product spaces: to make this more explicit we can write

${\displaystyle \psi _{\mathcal {B}}:(V,\phi )\rightarrow (\mathbb {R} ^{n},\delta _{ij}).}$

Explicitly we can write ${\displaystyle (\psi _{\mathcal {B}}(v))^{i}=e^{i}(v)=\phi (e_{i},v)}$ where ${\displaystyle e^{i}}$ is the dual basis element to ${\displaystyle e_{i}}$.

The inverse is a component map

${\displaystyle C_{\mathcal {B}}:\mathbb {R} ^{n}\rightarrow V,(v^{i})\mapsto \sum _{i=1}^{n}v^{i}e_{i}.}$

These definitions make it manifest that there is a bijection

${\displaystyle \{{\text{Space of orthogonal bases }}{\mathcal {B}}\}\leftrightarrow \{{\text{Space of isomorphisms }}V\leftrightarrow \mathbb {R} ^{n}\}.}$

The space of isomorphisms admits actions of orthogonal groups at either the ${\displaystyle V}$ side or the ${\displaystyle \mathbb {R} ^{n}}$ side. For concreteness we fix the isomorphisms to point in the direction ${\displaystyle \mathbb {R} ^{n}\rightarrow V}$, and consider the space of such maps, ${\displaystyle {\text{Iso}}(\mathbb {R} ^{n}\rightarrow V)}$.

This space admits a left action by the group of isometries of ${\displaystyle V}$, that is, ${\displaystyle R\in {\text{GL}}(V)}$ such that ${\displaystyle \phi (\cdot ,\cdot )=\phi (R\cdot ,R\cdot )}$, with the action given by composition: ${\displaystyle R*C=R\circ C.}$

This space also admits a right action by the group of isometries of ${\displaystyle \mathbb {R} ^{n}}$, that is, ${\displaystyle R_{ij}\in {\text{O}}(n)\subset {\text{Mat}}_{n\times n}(\mathbb {R} )}$, with the action again given by composition: ${\displaystyle C*R_{ij}=C\circ R_{ij}}$.

The set of orthonormal bases for ${\displaystyle \mathbb {R} ^{n}}$ with the standard inner product is a principal homogeneous space or G-torsor for the orthogonal group ${\displaystyle G={\text{O}}(n),}$ and is called the Stiefel manifold ${\displaystyle V_{n}(\mathbb {R} ^{n})}$ of orthonormal ${\displaystyle n}$-frames.[7]

In other words, the space of orthonormal bases is like the orthogonal group, but without a choice of base point: given the space of orthonormal bases, there is no natural choice of orthonormal basis, but once one is given one, there is a one-to-one correspondence between bases and the orthogonal group. Concretely, a linear map is determined by where it sends a given basis: just as an invertible map can take any basis to any other basis, an orthogonal map can take any orthogonal basis to any other orthogonal basis.

The other Stiefel manifolds ${\displaystyle V_{k}(\mathbb {R} ^{n})}$ for ${\displaystyle k<n}$ of incomplete orthonormal bases (orthonormal ${\displaystyle k}$-frames) are still homogeneous spaces for the orthogonal group, but not principal homogeneous spaces: any ${\displaystyle k}$-frame can be taken to any other ${\displaystyle k}$-frame by an orthogonal map, but this map is not uniquely determined.

• The set of orthonormal bases for ${\displaystyle \mathbb {R} ^{p,q}}$ is a G-torsor for ${\displaystyle G={\text{O}}(p,q)}$.
• The set of orthonormal bases for ${\displaystyle \mathbb {C} ^{n}}$ is a G-torsor for ${\displaystyle G={\text{U}}(n)}$.
• The set of orthonormal bases for ${\displaystyle \mathbb {C} ^{p,q}}$ is a G-torsor for ${\displaystyle G={\text{U}}(p,q)}$.
• The set of right-handed orthonormal bases for ${\displaystyle \mathbb {R} ^{n}}$ is a G-torsor for ${\displaystyle G={\text{SO}}(n)}$

• Orthogonal basis
• Basis (linear algebra) – Set of vectors used to define coordinates
• Orthonormal frame – Euclidean space without distance and angles
• Schauder basis – Computational tool
• Total set

• Roman, Stephen (2008). Advanced Linear Algebra. Graduate Texts in Mathematics (Third ed.). Springer. ISBN 978-0-387-72828-5. (page 218, ch.9)
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.

• This Stack Exchange Post discusses why the set of Dirac Delta functions is not a basis of L²([0,1]).