Let $M$ be a finite dimensional subspace of a complex Hilbert space $H$, and suppose that $H$ has an orthonormal basis $\{e_k:k\in\mathbb{N}\}$. Let $P$ denote the orthogonal projection from $H$ onto $M$. Prove that $\dim{M} = \sum_{k=1}^\infty ||P(e_k)||^2$, where $||\cdot||$ denotes the natural norm.
So far, I have found an orthonormal basis $\{f_1,…,f_n\}$ for $M$ and then used that $P(x)=\sum_{k=1}^n \langle x,f_k \rangle f_k$ for any $x \in H$. Then I get stuck when trying to find $P(e_k)$ and simplify things. Any tips please?
Best Answer
You have that $$ \|P_M x\|^2=\sum_{j=1 }^n|\langle x ,f_j\rangle|^2 $$ for any orthonormal basis $f_1,\ldots ,f_n$ of $M$ and any chosen $x\in H$. Hence $$ \sum_{k\in \Gamma }\|P_M(e_k)\|^2=\sum_{k\in \Gamma }\sum_{j=1}^n|\langle e_k,f_j \rangle|^2\overset{(*)}{=}\sum_{j=1}^n\sum_{k\in \Gamma }|\langle e_k,f_j \rangle|^2\overset{(**)}{=}\sum_{j=1}^n\|f_j\|^2=n $$ where in $(*)$ we used Tonelli's theorem together with the definition of convergence of an unordered sum of positive elements, and in $(**)$ we used Parseval's identity.
That is: if the unordered sum $\sum_{k\in \Gamma }a_k$ converges and $a_k\geqslant 0$ for all $k\in \Gamma $, then it can be shown that the set $\{k\in \Gamma :a_k\neq 0\}$ is countable.
In your case the solution is simpler because $\Gamma =\Bbb N $, so you dont need to know about convergence of unordered sums.