proof of $\langle z,w\rangle =\overline{\langle w,z\rangle}$

Question

Here is an property of inner product space V: $\langle z,w\rangle =\overline{\langle w,z\rangle}$ (conjugate symmetry).

I need a reference on the proof of this property or general idea in proving this?

Answer 1

As pointed out in the comments, the relation

$\langle z, w \rangle = \overline{\langle w, z \rangle } \tag 1$

is usually taken as an axiom; however, it is motivated by, and derives from, the corresponding property for the standard (hermitian) inner product on $\Bbb C^n$:

$\langle z, w \rangle = \displaystyle \sum_1^n \bar z_j w_j; \tag 2$

for this inner product we have

$\overline{\langle w, z \rangle} = \overline{ \displaystyle \sum_1^n \bar w_j z_j } = \sum_1^n \bar{\bar w_j} \bar z_j = \sum_1^n w_j \bar z_j = \langle z, w \rangle. \tag 3$

The axiom (1) abstracts this to more general contexts in which $\Bbb C^n$ may not be directly available.

Also, see the comment by Arturo Magidin below.

Answer 2

Suppose $w, z \in \mathbb{C}^n$. Using the additive properties of complex conjugation, $$ \langle z, w \rangle = \sum_{i = 1}^{n} z_i \overline{w}_i = \overline{\overline{\sum_{i = 1}^{n} z_i \overline{w}_i}} = \overline{\sum_{i = 1}^n \overline{z_i \overline{w}_i}} = \overline{\sum_{i = 1}^{n} \overline{z}_i w_i} = \overline{\sum_{i = 1}^{n} w_i \overline{z}_i} = \overline{\langle w, z \rangle} $$