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"?

 A: Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
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.
