Dense Countable basis on Hilbert space

Question

Let say that I have a $H$ hilbert space and linear independent countable set $\beta =\{ \beta_1 , \beta_2, \beta_3... \}$

such that $span(\beta)$ is dense set in H.

does $span(\beta-\beta_1) =span( \{ \beta_2, \beta_3... \})$

is dense set in $H$ ??

Answer 1

There are certainly instances where this is false. As Jochen said, if this is an orthonormal basis, then $span(\beta-\beta_1) \subset \{\beta_1\}^\perp$ and so is not dense. There are bases where it is possible to remove one or any finite number of terms and keep density, these are called over complete bases.

For instance in a RKHS, the span of the kernel functions are often overcomplete.

A more elementary example is bases as in the following:

If $\{\beta_n\}_{n=-\infty}^\infty$ is an orthonormal basis then the span of $\{\beta_n - \beta_{n-1}\}_{n=-\infty}^\infty$ is dense in the Hilbert space and continues to span the space if any one element is removed. This is from Halmos's "A Hilbert Space Problems Book".

Answer 2

If, for example, $\beta_n$ are orthogonal then span$\lbrace \beta_2,\beta_3,\ldots\rbrace$ is not dense. The distance of $\beta_1$ to this set is $\|\beta_1\|$.