Does an Orthogonal Projection map a basis of the space to a spanning set of the its subspace for a Hilbert Space?

Question

For finite dimensional vector spaces it is quite easy to prove that a surjective linear operator $V\mapsto W$ maps a basis of $V$ to a spanning set of $W$. Is this property still true for linear operators on Hilbert spaces? Specifically, I'm interested in the case of orthogonal projections and when the Hilbert space is separable.

Answer 1

Let $\{e_i\}$ be a basis for $V$. Let $W$ be a subspace of $V$, and $p: V \rightarrow W$ a projection, i.e. a surjective linear map such that $p^2 = p$.

Every element $w$ of $W$ is expressible as $\sum\limits_i c_i e_i$ for unique scalars $c_i$, almost all zero. Then

$$w = p(w) = \sum\limits_i c_i p(e_i).$$

This shows that $\{ p(e_i)\}$ spans $W$.

Similar statement for separable Hilbert spaces:

Let $e_1, e_2, e_3, ...$ be an orthonormal basis for $V$. Let $W$ be a closed subspace of $V$, and let $p: V \rightarrow W$ be a bounded linear operator such that $p = p^2$. For every element $w \in W$, there exist unique scalars $c_i$ such that in the norm topology on $V$,

$$w = \lim\limits_{n \to \infty} \sum\limits_{i=1}^nc_i e_i.$$

Then also $w = p(w) = \lim\limits_{n \to \infty} \sum\limits_{i=1}^nc_i p(e_i)$, this being a norm limit in $W$.