Math 2.0

Functional analysis Theorem

Riesz representation theorem in finite dimensions

Builds on Inner product spaceLinear form

Statement

Let VV be a finite-dimensional inner product space over ℝ\R or ℂ\C, with the inner product linear in the first argument. Then, for each linear form f ⁣:V→𝕂f \colon V \to \mathbb{K}, there is exactly one vector y∈Vy \in V with f(x)=⟨x,y⟩for all x∈V.f(x) = \langle x, y \rangle \qquad \text{for all } x \in V.

Remarks

For an orthonormal basis e1,…,ene_1, \dots, e_n, the vector is y=∑if(ei)‾ eiy = \sum_i \overline{f(e_i)}\, e_i. The map y↦⟨⋅,y⟩y \mapsto \langle \cdot, y \rangle from VV to its dual is bijective; it is linear over ℝ\R and conjugate linear over ℂ\C. The statement holds for continuous linear forms on every Hilbert space.

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Vector space, definition, no evidence levelVector spacedefinitionInner product, definition, no evidence levelInner productdefinitionLinear map, definition, no evidence levelLinear mapdefinitionInner product space, definition, no evidence levelInner product spacedefinitionLinear form, definition, no evidence levelLinear formdefinitionRiesz representation theorem in finite dimensions, statement, no evidence levelRiesz representation …statement
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  1. Vector space definition
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  5. Linear form definition
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Further reading

  1. Web https://en.wikipedia.org/wiki/Riesz_representation_theorem, Example in finite dimensions using matrix transformations
About this page
Tag04GYPXDXNH
URIm2://04GYPXDXNH
Kindstatement, theorem
WritingW0 stub
MSC46C05, 15A04
StatusThis copy of the site has no build data. Run m2 status 04GYPXDXNH.
UsesInner product space, Linear form

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