Let $\left\{e_1, \ldots, e_n\right\}$ be a finite orthonormal system in an inner product space $(E,\langle\cdot, \cdot\rangle)$ and let us abbreviate $F:=\operatorname{span}\left\{e_1, \ldots, e_n\right\}$. The mapping
$$
P: E \longrightarrow E, \quad P f=\sum_{j=1}^n\left\langle f, e_j\right\rangle e_j
$$
is called the orthogonal projection onto the subspace $F$. I would like to prove that $P$ does only depend on the subspace $F$ and not on the chosen orthonormal basis of $F$ used in the construction of $P$.
My attempt
Let $f, g \in E$ such that $g \in F$ and $f-g \perp F$, I have to prove that $g=P f$. Since $f-g \perp F$ and $f-Pf\perp F$, we have that
$$\langle f-g, e_i\rangle=\langle f-Pf, e_i\rangle=0\quad \forall i$$
then
$$\langle f-g, e_i\rangle-\langle f-Pf, e_i\rangle=0$$
and so
$$\langle f-g -(f-Pf), e_i\rangle=0\quad \forall i$$
from which
$$\langle Pf-g, e_i\rangle=0\quad \forall i$$
And now? How could I proceed? I think that $Pf-g\perp F$ is not true, so we should have $Pf=g$.
Best Answer
A direct approach might be better suited for the problem.
Let $h_1, \dots , h_n$ be another orthonormal basis of $F$ and $P^\prime$ the projection defined using it. Then using the sesquilinearity of $\langle \cdot , \cdot \rangle$ and that both $h_1, \dots , h_n$ and $e_1, \dots , e_n$ form an orthonormal basis of $F$: \begin{align*} Pf &= \sum_i \langle P f , h_i \rangle h_i = \sum_i \langle \sum_j \langle f, e_j\rangle e_j , h_i \rangle h_i\\ &= \sum_i \sum_j\langle f,e_j \rangle \langle e_j, h_i \rangle h_i = \sum_i \langle f, \sum_j\langle h_i, e_j \rangle e_j \rangle h_i = \sum_i \langle f, h_i \rangle h_i =P^\prime f \end{align*} for all $f \in E$.