Dual Vector Spaces with Orthonormal Basis I'm really stuck on the following quesiton.
Let $U$ and $V$ be finite dimensional vector spaces over the complex numbers with bases $e_1,..,e_n$ of $U$ and $f_1,...,f_m$ of $V$. They also have dual spaces $U^*$ and $V^*$ with bases $e^i$ and $f^i$ respectively.
Then assume that both spaces are Hermitian.
Let $T_U:U\to U^*$ be defined by $T_U(w)(u) = \langle w,u\rangle \forall w,u\in U$. 
Prove that if the basis $e_i$ is orthonormal then $T_U(e_i) = e^i$.
I have tried showing that
$T_U(e_i)(u) = \langle e_i, u\rangle $
$ = \langle e_i,x_1e_1+...+x_ne_n\rangle $
$ = \langle e_i, x_1e_1\rangle + \langle e_i, x_2e_2\rangle +\cdots +\langle e_i, x_ne_n\rangle$
$ =x_1\langle e_i,e_1\rangle + x_2\langle e_i,e_2\rangle +\cdots +x_n\langle e_i,e_n\rangle$
So we are left with, $x_i\langle e_i,e_i \rangle $ since the others are mutually orthogonal.
I'm having trouble understanding why this implies the desired result.
 A: The property of the dual basis is that $e^{i}(e_{j}) = \delta_{i,j}$. So, why don't you consider how $T_{U}(e_{i})$ and $e^{i}$ act on the basis of $U$?
A: To see that
$T_U(e_i) = e^i, \tag{1}$
first compute
$T_U(e_i)(e_j) = \langle e_i, e_j \rangle = \delta_{ij} \tag{2}$
by orthonormality of the $e_i$.  Then note that
$e^i(e_j) = \delta_{ij} \tag{3}$
as well, this time by the duality if the bases $e_i$, $e^j$.  Then since $T_U(e_i)$ and $e^i$ agree on the basis elements $e_j$ of $U$, it follows by linearity thet agree on every $u = \sum u_j e_j \in U$:
$T_U(e_i)(\sum u_j e_j) = \sum u_j T_U(e_i)(e_j) = \sum u_j e^i(e_j) = e^i(\sum u_j e_j); \tag{4}$
thus
$T_U(e_i) = e^i, \tag{5}$
and we are done!QED
Hope this helps.  Holiday Cheers,
and as always,
Fiat Lux!!!
A: Hint: First of all you don't need $V$ for your question, right? 
So let's first talk about your Hermitian. If $U$ is a Hermitian space, then there is a Hermitian form 
$$\langle, \rangle: U \times U \longrightarrow \Bbb C$$
Now your map $T_U$ is not defined in a clear way. It should go from $U \to U^*$, but then, you have to define
$$T_U: U \to U^*, u \mapsto \langle -,u \rangle$$
and then $T_U(u) \in U^*$ for $u \in U$. So to prove that $T_U(e_i) = e^i$, you just have to know how the $e^i$ is defined, namely via the Kronecker Delta:
$$e^i (e_j) = \delta_{ij}$$
So start with $u = \sum_{i=1}^{n} e_i$ and show what happens if you let $T_U$ act on $u$.
