# 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.

$\,$

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 $x\in H$

 $\|x\|^{2}=\sum_{i\in I}|\langle x,e_{i}\rangle|^{2}$

We know that, in the above sum, $\langle x,e_{i}\rangle\neq 0$ for only a countable number of $i\in I$. Thus, considering $x$ as $f_{j}$, the set $I_{j}:=\{i\in I:\langle f_{j},e_{i}\rangle\neq 0\}$ is countable. Since for each $i\in I$ we also have

 $\|e_{i}\|^{2}=\sum_{j\in J}|\langle e_{i},f_{j}\rangle|^{2}$

there must be $j\in J$ such that $\langle f_{j},e_{i}\rangle\neq 0$. We conclude that $\displaystyle I=\bigcup_{j\in J}I_{j}$.

Hence, since each $I_{j}$ is countable, $I\leq J\!\times\!\mathbb{N}\cong J$ (because $J$ is infinite).

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

Title all orthonormal bases have the same cardinality AllOrthonormalBasesHaveTheSameCardinality 2013-03-22 17:56:10 2013-03-22 17:56:10 asteroid (17536) asteroid (17536) 7 asteroid (17536) Theorem msc 46C05 dimension of an Hilbert space is well-defined