Exercise 2.6

Consider an inner product \((\cdot, \cdot)\) and we should have

(1)\[ \left(|v\rangle, \sum_{i}\lambda _i |w_i\rangle\right) = \sum_i \lambda_i (|v\rangle, |w\rangle) \]

Then using \((|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*\), we can compute

(2)\[ \left(\sum_{i}\lambda _i |w_i\rangle, |v\rangle\right) =\left(|v\rangle, \sum_{i}\lambda _i |w_i\rangle\right)^* =\sum_i \lambda^*_i (|v\rangle, |w\rangle)^* = \sum_i \lambda^*_i (|w\rangle, |v\rangle) \]

where we use relation \(\left(\sum_{i} z_i\right)^* = \sum_i z^*_i\) and \((z_1z_2)^* = z^*_1 z^*_2\). Therefore, we can conclude that any inner product \((\cdot, \cdot)\) is conjugate-linear in the first argument.