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Every finite-dimensional real or complex inner product space has an orthonormal basis
Statement
Every finite-dimensional real or complex inner product space has an orthonormal basis. In dimension zero, this is the empty basis.
Facts & Assumptions
Given: A finite-dimensional inner product space .
A finite-dimensional vector space has a finite basis, with the empty list serving when (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).
Gram–Schmidt converts every finite independent list into an orthonormal list with the same span (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).
Proof
Choose a finite basis of using [L1].
Apply [L2]. The resulting orthonormal list has the same span as the basis, namely , and therefore is an orthonormal basis. This also covers .
Depends on
Used by
- Orthogonal and unitary operators form groups, and their determinants have modulus one Corollary
- Finite-dimensional Riesz representation: every functional is uniquely v↦⟨ v,w⟩ Theorem
- For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and T^*T=I are equivalent Theorem
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 53 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Sheldon Axler, Linear Algebra Done Right, 4th ed., result 6.35 (standard reference, not scraped)