Let
- $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$
- $U$ and $H$ be $\mathbb K$-Hilbert spaces
- $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$
- $\iota:U\to H$ be an embedding and $V:=\iota(U)$
- $Q$ denote the orthogonal projection of $H$ onto $V$
Can we build an orthonormal basis of $V$ from $(Qe_n)_{n\in\mathbb N}$?
Say $x \in V$. Since the $e_n$ are a basis for $H$ and $x \in H$, $x = \sum_n c_n e_n$ for some coefficients $c_n \in \mathbb{K}$. Since $Q$ and $1-Q$ are orthogonal projections whose sum is 1, $x = Qx + (1-Q)x = (Q \sum_n c_n e_n) + ((1-Q) \sum_n c_n e_n)$. But the second term is zero since $x \in V$ implies $Qx = x$. So we have $x = Q \sum_n c_n e_n$, and by linearity, $x = \sum_n c_n Q e_n$. Thus any $x \in V$ can be written as a linear combination of $Q e_n$'s, and $(Q e_n)_{n \in \mathbb{N}}$ is therefore a basis for $V$. However it doesn't have to be an orthogonal basis, or even a linearly independent one. For example, $V$ could simply be multiples (real or complex depending on $\mathbb{K}$) of $e_1 + e_2$, say, in which case $Q e_1 = Q e_2 = (e_1 + e_2) / \sqrt{2}$. So just run the Gram-Schmidt procedure on the $Q e_n$, throwing out any linearly dependent ones you come across, and you should end up with an orthonormal basis.