The linear dimension of a vector space $V$ over a field $F$ is the cardinality of a linearly independent set that spans the vector space. If $V$ is an inner product space then it acquires another definition of orthonormal dimension which is the cardinality of a complete orthonormal set in $V$. For finite-dimensional Vector Spaces, both linear and orthonormal dimensions are equal. I need an example of an inner product space $V$ for which the orthonormal dimension is not equal to linear dimension, in case such a $V$ exists.
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Thank you for the alert. I have edited my question. – Lawrence Mano Sep 22 '20 at 11:07
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Complete orthonormal set is one which is not properly contained in another orthonormal set and having largest cardinality. – Lawrence Mano Sep 22 '20 at 11:12
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Relevant https://math.stackexchange.com/questions/13641/an-orthonormal-set-cannot-be-a-basis-in-an-infinite-dimension-vector-space?rq=1 – Sumanta Sep 22 '20 at 11:14
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In every separable Hilbert space $H$ there is countable orthonormal basis. So therefore the orthogonal dimmension is equal to $\aleph_0$. If we assume that the algebraical dimmension is also $\aleph_0$ then there should exist an algebraical basis $\{e_k : k\in \mathbb{N}\} $ of $H.$ Consider $$H_n =\mbox{span} \{ e_1 , e_2 , ..., e_n\}$$ the subspace spanned by the vectors $ \{ e_1 , e_2 , ..., e_n\}.$ The subspaces $H_n $ are closed since they are finite dimensional. Moreover $$H=\bigcup_{n=1}^{\infty } H_n $$ therefore since $H$ is a complete metric space we get by the Theorem of Baire that there exist a open ball $B$ of $H$ and $n\in\mathbb{N} $ such that $B\subset H_n.$ But this implies that $H\subset H_n $ which is imposible. Therefore the algebraical dimmension of $H$ must be greater than $\aleph_0.$
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If you take the space $L^2[0,1]$ inner proudct space, than it has a countable orthonormal dimesnion, with orthonormal basis $\{e^{2\pi m i} \}_{m\in \mathbb{Z}}$. But its Hamel dimension is uncountable. In fact if $\vert V\vert > \aleph$, then its Hamel dimension would be equal to $\vert V\vert$.
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