# Riesz, Hilbert and Hamel bases

I was surprised to read both at PlanetMath and in Wikipedia (apparently copied from PlanetMath) that

If $H$ is a finite-dimensional [Hilbert] space, then every basis of $H$ is a Riesz basis.

I thought that in the context of Hilbert spaces the unqualified term "basis" is usually taken to mean "Hilbert basis" (i.e. orthonormal basis), not "Hamel basis" (i.e. basis in the linear-algebra sense), but under that interpretation the restriction to finite-dimensional spaces makes no sense, since every Hilbert basis is a Riesz basis (with constants $c=C=1$). Am I right in thinking that

• they mean a Hamel basis, and
• it would be preferable to disambiguate that statement by inserting "Hamel"?
• On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space... – user53153 Dec 19 '12 at 0:32

On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $\mathbb{Q}$-vector space $\mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.