orthogonal basis

Orthogonal and orthonormal subsets


For VV an inner product space, a collection of vectors {v i}\{v_i\} of VV is orthogonal with respect to the inner ,:V×Vk\langle{-, -}\rangle\colon V \times V \to k if for all iji \neq j we have v i,v j=0\langle{v_i, v_j}\rangle = 0. It is orthonormal if additionally we have v i,v i=1\langle{v_i, v_i}\rangle = 1; that is, v i,v j=δ i,j\langle{v_i, v_j}\rangle = \delta_{i,j} (the Kronecker delta) for all i,ji, 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 VV, due attention must be paid to the meaning of ‘basis’. When VV has finite dimension, one can just use the purely algebraic sense of basis: a linearly independent set whose span is all of VV, or equivalently a maximal linearly independent set. In this case, there is no restriction on the ground field kk.

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

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 v,v0\langle v, v \rangle \neq 0 for v0v \neq 0 (as for example in the case of a positive definite inner product space), and assuming square roots v,v 1/2\langle v, v \rangle^{1/2} exist in the ground field kk, 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.

Revised on May 25, 2017 11:26:30 by Urs Schreiber (