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.