Orthogonal representation of finite operator I would like to know if my proof is correct.
Statement: Let $T$ be a finite rank operator on a Hilbert space $\mathscr{H}$. Show that $\forall \, h \, \in \mathscr{H}, \, T(h)$ can be written as $T(h) = \sum_{i=1}^n\langle h, e_i \rangle f_i$ where $\{e_i\}$ is an orthonormal basis of $\mathscr{H}$ 
Attempted Proof: $T(h)$ can be written as $T(h) = \sum_{i=1}^n\langle T(h), e_i \rangle e_i = \sum_{i=1}^n\langle h, T^*(e_i) \rangle e_i$.
Now, if we let $f_i=T^*(e_i)$ then:
$T(h) = \sum_{i=1}^n\langle T(h), e_i \rangle e_i = \sum_{i=1}^n\langle h, T^*(e_i) \rangle e_i = \sum_{i=1}^n\langle h, f_i \rangle e_i$ 
And this is where I get stuck.
 A: I don't see the point of your last computation: you have already obtained what you are looking for before the "And if we take..."
You got your $e_i$ and $f_i$ in the wrong spots, but that's just a labeling issue: you should have started with $\{f_i\}$ as the orthonormal basis. 
On a deeper side, why would $\{T^*e_i\}$ and orthonormal basis? In general, it isn't. 
A: Let $\{ e_{j}\}_{j=1}^{N}$ be an orthonormal basis of $\mathcal{R}(T^{\star}T)$--which is finite-dimensional and, hence, closed--and extend this to a full orthonormal basis of $X$ by adding an orthonormal basis $\{ e_{j} \}_{j=N+1}^{\infty}$ of $\mathcal{R}(T^{\star}T)^{\perp}=\mathcal{N}(T^{\star}T)=\mathcal{N}(T)$. Then $Te_{j}=0$ for $j > N$, and
$$
        Tx = T\sum_{j=1}^{\infty}(x,e_{j})e_{j}=\sum_{j=1}^{N}(x,e_{j})Te_{j}=\sum_{j=1}^{N}(x,e_{j})f_{j}.
$$
Note: I am assuming that $\mathscr{H}$ is separable, but this seems reasonable based on the fact that you have suggested that $\mathscr{H}$ has a countable orthonormal basis. However, the final sum representation $Tx = \sum_{j=1}^{N}(x,e_{j})f_{j}$ does not really depend on such an assumption because $T$ annihilates $\mathcal{N}(T^{\star}T)=\mathcal{N}(T)$, regardless of its dimension.
