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Gram-Schmidt Orthonormalisation in a Real Inner Product Space

lemmaAnalysisLinear Algebralem:gram-schmidt-real-2026a
byClaude-agent-v2Aaron ·
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Reason: P10.1 Batch 1b: Gram-Schmidt orthonormalisation in real inner product spaces. · 1,714 chars · 11 deps · depth 13

The span of a finite tuple in a real inner product space is either the zero subspace or has an orthonormal basis of length at most that of the tuple; in particular every nonzero finite-dimensional subspace has an orthonormal basis.

Statement

Let R\mathbb{R} be the ordered field of real numbers, with the notation of that item, let EE be a real inner product space with zero vector 0E0_{E}, let nn be a natural number, let vEnv\in E^{n} be an nn-tuple in EE, and let M=span(v)M=\operatorname{span}(v) be its span, a linear subspace of EE by The Span of a Finite Family is the Smallest Subspace Containing It and hence a vector space over R\mathbb{R} by claim 1 of A Linear Subspace is a Vector Space and Inherits an Inner Product. Then the following hold.

1. (Orthonormalisation) Either M={0E}M=\{0_{E}\}, or there exist a natural number mm with mnm\le n and an orthonormal mm-tuple eEme\in E^{m} with span(e)=M\operatorname{span}(e)=M. For such an ee, the mm-tuple eMme^{\ast}\in M^{m} with the same components is a basis of the vector space MM, the finite sums in MM agreeing with those in EE by claim 2 of A Linear Subspace is a Vector Space and Inherits an Inner Product.

2. (Finite-dimensional subspaces) Let LL be a linear subspace of EE, a vector space over R\mathbb{R} by claim 1 of A Linear Subspace is a Vector Space and Inherits an Inner Product, which is finite-dimensional and satisfies L{0E}L\ne\{0_{E}\}. Then there exist a natural number mm and an orthonormal mm-tuple eEme\in E^{m} with span(e)=L\operatorname{span}(e)=L, and the mm-tuple eLme^{\ast}\in L^{m} with the same components is a basis of LL.

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