nLab orthogonal basis

Definition

For $V$ an inner product space, a collection of vectors $\{v_i\}$ of $V$ is orthogonal with respect to the inner $\langle{-, -}\rangle\colon V \times V \to k$ if for all $i \neq j$ we have $\langle{v_i, v_j}\rangle = 0$. It is orthonormal if additionally we have $\langle{v_i, v_i}\rangle = 1$; that is, $\langle{v_i, v_j}\rangle = \delta_{i,j}$ (the Kronecker delta) for all $i, j$. Note that an orthogonal set is necessarily linearly independent, assuming (as one often does) that the inner product is nondegenerate.

Bases

In speaking of the notion of orthonormal or orthogonal basis of $V$, due attention must be paid to the meaning of ‘basis’. When $V$ has finite dimension, one can just use the purely algebraic sense of basis: a linearly independent set whose span is all of $V$, or equivalently a maximal linearly independent set. In this case, there is no restriction on the ground field $k$.

Often, however, ‘basis’ is supposed to mean the sense of basis used for a topological vector space $V$, i.e., a linearly independent set whose span is dense in $V$. In the TVS case, it is usually understood that the ground field $k$ is a local field, and the inner product is separately continuous with respect to the topology on $V$. A typical situation is where $k$ is $\mathbb{R}$ or $\mathbb{C}$ and the inner product is a positive definite inner product on $V$ which naturally induces a metric and from there a TVS structure on $V$.

We have this interesting pair of results:

• On one hand, no infinite-dimensional positive-definite inner-product space has an orthogonal (much less orthonormal) basis in the purely algebraic sense.
• On the other hand, any maximal orthogonal set (which must exist by an application of Zorn's lemma) is necessarily a basis in the topological sense.

Gram–Schmidt process

For a finite-dimensional vector space equipped with a nondegenerate inner product, one may replace any well-ordered basis by an well-ordered orthogonal basis using the Gram–Schmidt process. Assuming that we have $\langle v, v \rangle \neq 0$ for $v \neq 0$ (as for example in the case of a positive definite inner product space), and assuming square roots $\langle v, v \rangle^{1/2}$ exist in the ground field $k$, we can further produce a well-ordered orthonormal basis from a well-ordered basis.

In the TVS case, with the appropriate meaning of basis understood, the Gram-Schmidt process extends without difficulty to any well-ordered basis, because the process yields a linearly independent orthogonal set with the same span as the original basis.

Examples

• Schur's lemma says that irreducible representations form an orthogonal basis of the representation ring. See there for more.

Related concepts

• frame field

• orthogonality

• orthogonal structure

• normal vector

• mutually unbiased bases

References

See also

• Wikipedia, Orthogonal basis