proof of Gram-Schmidt orthogonalization procedure

Note that, while we state the following as a theoremMathworldPlanetmath for the sake of logical completeness and to establish notation, our definition of Gram-Schmidt orthogonalizationPlanetmathPlanetmath is wholly equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to that given in the defining entry.


(Gram-Schmidt Orthogonalization) Let {uk}k=1n be a basis for an inner product spaceMathworldPlanetmath V with inner productMathworldPlanetmath <,>. Define v1=u1u1 and {mk}k=2n recursively by

𝐦𝐤=𝐮𝐤-<𝐮𝐤,𝐯𝟏>𝐯𝟏-<𝐮𝐤,𝐯𝟐>𝐯𝟐--<𝐮𝐤,𝐯𝐤-𝟏>𝐯𝐤-𝟏, (1)

where vk=mkmk for 2kn. Then {vk}k=1n is an orthonormal basisMathworldPlanetmath for V.


We proceed by inductionMathworldPlanetmath on n. In the case n=1, we suppose {𝐮𝐤}k=1n={𝐮𝐤}k=11=𝐮𝟏 is a basis for the inner product space V. Letting 𝐯𝟏=𝐮𝟏𝐮𝟏, it is clear that 𝐯𝟏Span(𝐮𝟏), whence it follows that Span(𝐯𝟏)=Span(𝐮𝟏)=V. Thus 𝐯𝟏 is an orthonormal basis for V, and the result holds for n=1. Now let n1, and suppose the result holds for arbitrary n. Let {𝐮𝐤}k=1n+1 be a basis for an inner product space V. By the inductive hypothesis we may use {𝐮𝐤}k=1n to construct an orthonormal setMathworldPlanetmath of vectors {𝐯𝐤}k=1n such that Span({𝐯𝐤}k=1n)=Span({𝐮𝐤}k=1n). In accordance with the procedure outlined in the statement of the theorem, let 𝐦𝐧+𝟏 be defined as


First we show that the vectors 𝐯𝟏,𝐯𝟐,,𝐯𝐧,𝐦𝐧+𝟏 are mutually orthogonalMathworldPlanetmathPlanetmath. Consider the inner product of 𝐦𝐧+𝟏 with 𝐯𝐣 for 1jn. By construction, we have


Now since {𝐯𝐤}k=1n is an orthonormal set of vectors, whence <𝐯𝐢,𝐯𝐣δij, each term in the summation on the right-hand side of the preceding equation will vanish except for the term where i=j. Thus by this and the preceding equation, we have


Thus 𝐦𝐧+𝟏 is orthogonal to 𝐯𝐣 for 1jn, so we may take 𝐯𝐧+𝟏=𝐦𝐧+𝟏𝐦𝐧+𝟏 to have {𝐯𝐤}k=1n+1 an orthonormal set of vectors. Finally we show that {𝐯𝐤}k=1n+1 is a basis for V. By construction, each 𝐯𝐤 is a linear combinationMathworldPlanetmath of the vectors {𝐮𝐤}k=1n+1, so we have n+1 orthogonal, hence linearly independentMathworldPlanetmath vectors in the n+1 dimensional space V, from which it follows that {𝐯𝐤}k=1n+1 is a basis for V. Thus the result holds for n+1, and by the principle of induction, for all n. ∎

Title proof of Gram-Schmidt orthogonalization procedure
Canonical name ProofOfGramSchmidtOrthogonalizationProcedure
Date of creation 2013-03-22 16:27:17
Last modified on 2013-03-22 16:27:17
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 11
Author rspuzio (6075)
Entry type Proof
Classification msc 65F25
Synonym Gram-Schmidt
Synonym orthogonalization
Related topic GramSchmidtOrthogonalization
Related topic ExampleOfGramSchmidtOrthogonalization
Related topic InnerProductSpace
Related topic InnerProduct
Related topic QRDecomposition
Related topic NormedVectorSpace
Related topic Orthogonal
Related topic OrthogonalVectors
Related topic Basis
Related topic Span
Related topic LinearIndependence
Related topic Orthonormal
Related topic OrthonormalBasis
Defines Gram-Schmidt orthogonalization