all orthonormal bases have the same cardinality

Theorem. – All orthonormal bases of an Hilbert space $H$ have the same cardinality. It follows that the concept of dimension of a Hilbert space is well-defined.

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Proof: When $H$ is finite-dimensional (as a vector space), every orthonormal basis is a Hamel basis of $H$. Thus, the result follows from the fact that all Hamel bases of a vector space have the same cardinality (see this entry (http://planetmath.org/AllBasesForAVectorSpaceHaveTheSameCardinality)).

We now consider the case where $H$ is infinite-dimensional (as a vector space). Let ${\{e_i\}}_{i\in I}$ and ${\{f_j\}}_{j\in J}$ be two orthonormal basis of $H$, indexed by the sets $I$ and $J$, respectively. Since $H$ is infinite dimensional the sets $I$ and $J$ must be infinite.

We know, from Parseval’s equality, that for every $x\in H$

$${\parallel x\parallel }^2=\sum _{i\in I}{|⟨x,e_i⟩|}^2$$

We know that, in the above sum, $⟨x,e_i⟩\ne 0$ for only a countable number of $i\in I$. Thus, considering $x$ as $f_j$, the set $I_j:=\{i\in I:⟨f_j,e_i⟩\ne 0\}$ is countable. Since for each $i\in I$ we also have

$${\parallel e_i\parallel }^2=\sum _{j\in J}{|⟨e_i,f_j⟩|}^2$$

there must be $j\in J$ such that $⟨f_j,e_i⟩\ne 0$. We conclude that $I={\displaystyle \bigcup _{j\in J}}{I}_j$.

Hence, since each $I_j$ is countable, $I\le J\times \mathbb{N}\cong J$ (because $J$ is infinite).

An analogous proves that $J\le I$. Hence, by the Schroeder-Bernstein theorem $J$ and $I$ have the same cardinality. $\mathrm{\square }$

Title	all orthonormal bases have the same cardinality
Canonical name	AllOrthonormalBasesHaveTheSameCardinality
Date of creation	2013-03-22 17:56:10
Last modified on	2013-03-22 17:56:10
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	7
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46C05
Synonym	dimension of an Hilbert space is well-defined