Let $\langle x, y\rangle$ be an inner product on a vector space $V$, and let $e_1, e_2, \cdots, e_n$ be an orthonormal basis for $V$.
Question.
Prove: $\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = a_1b_1 + a_2b_2 + \cdots + a_nb_n$, where $a_i,b_i\in F$.
Answer attempt.
Since $\{e_1, e_2, \cdots, e_n\}$ is an orthonormal set, $\langle e_i, e_j\rangle = 0$ when $i \neq j$ and $\langle e_i, e_j\rangle = 1$ when $i = j$. So:
$\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = \langle a_1e_1, b_1e_1\rangle + \langle a_2e_2, b_2e_2\rangle + \cdots + \langle a_ne_n, b_ne_n\rangle = a_1\overline{b_1} + a_2\overline{b_2} + \cdots + a_n\overline{b_n}$
Associated question
Prove: $\langle x, y\rangle = \langle x, e_1\rangle\langle y, e_1\rangle + \langle x, e_2\rangle\langle y, e_2\rangle + \cdots + \langle x, e_n\rangle\langle y, e_n\rangle$
Answer attempt
Similarly, I get:
$\langle x, y\rangle = \langle x, e_1\rangle\overline{\langle y, e_1\rangle} + \langle x, e_2\rangle\overline{\langle y, e_2\rangle} + \cdots + \langle x, e_n\rangle\overline{\langle y, e_n\rangle}$
…am I missing something? If yes, I'd really appreciate it if someone would point out what.
Best Answer
The second follows immediately from the first, using $x=a_1e_1+\dots+a_ne_n$ and $y=b_1e_1+\dots+b_ne_n$.
There shouldn't be any complex conjugate symbols. You said inner product, not Hermitian inner product.