**linear algebra**, **higher linear algebra**

(…)

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.

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.

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.

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

See also

- Wikipedia,
*Orthogonal basis*

Last revised on June 7, 2023 at 07:25:54. See the history of this page for a list of all contributions to it.