Hilbert space basis which is not a vector space basis

Question

Is the set $(e_n)_{n>0}$ a (vector space) basis for the sequence Hilbert space $l^2$? It is a Hilbert space basis anyway.

I would say no, because the sequence $\left(\frac{1}{n}\right)_{n>0}$ is in $l^2$ but it can't be written as a finite linear combination of $e_i$'s.

Is that right?

Answer 1

Yes, it is entirely correct. Actually, it can be proved that a Hilbert space is infinite-dimensional if and only if no Hilbert basis is a basis in the Linear Algebra sense.

Answer 2

Yes, that's true. Any actual vector space basis for $\ell^2$ has to have the same size as $\mathbb{R}$ and cannot be explicitly written down.