Dimension of a closed linear subspace of a separable Hilbert space

Question

Given a separable Hilbert space $(X,(\cdot,\cdot))$, a closed linear subspace $V\subset X$ and an orthonormal basis $\{e_i\ |\ i\in\mathbb{N}\}$ for $X$, show that \begin{equation} \dim V = \sum_{i=1}^\infty \|Pe_i\|^2, \end{equation}

where $\|\cdot\|: X \to \mathbb{R} : x \mapsto \sqrt{(x,x)}$ and $P$ is the projection of $X$ onto $V$.

The steps I have taken:

1. I have shown that for all elements $x_1,x_2 \in X$, it holds that $(Px_1,x_2) = (x_1, Px_2)$.
2. If $\{v_i\ |\ i \in I \subset \mathbb{N}\}$ is a basis for $V$ with $|v_i|=1$ for each $i$, then \begin{equation} \dim V = \sum_{i\in I}\|v_i\|^2. \end{equation}
3. By the fact that $X$ is separable and that $\{e_i\ |\ i\in\mathbb{N}\}$ is an orthonormal basis for $X$, there exists a countable dense subsets $E\in X$ such that \begin{equation} \overline{\text{span}}\ E = X = \overline{\text{span}}\ \{e_i\ |\ i\in\mathbb{N}\}. \end{equation}

From this point, I don't know how to connect this to obtain a proof. Any suggestions? Thanks in advance!

Answer 1

If $(f_j)_{j \in J}$ is an orthonormal basis for $V$, with $\dim V = \operatorname{card} J$, then the orthogonal projection $P : X \to V$ is given by

$$Px = \sum_{j \in J} \langle x, f_j\rangle f_j$$

Then in particular $$Pe_i = \sum_{j \in J} \langle e_i, f_j\rangle f_j \implies \|Pe_i\|^2 = \sum_{j\in J} \left|\langle e_i, f_j\rangle \right|^2$$

So we have

$$\sum_{i=1}^\infty \|Pe_i\|^2 = \sum_{i=1}^\infty \sum_{j\in J} \left|\langle e_i, f_j\rangle \right|^2 = \sum_{j \in J} \sum_{i=1}^\infty \left|\langle e_i, f_j\rangle \right|^2 \stackrel{\text{Parseval}}{=} \sum_{j \in J} \|f_j\|^2 = \sum_{j \in J} 1 = \operatorname{card} J = \dim V$$

We can exchange the sums because $\left|\langle e_i, f_j\rangle \right|^2 \ge 0$.