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

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?

• This is an axiom for inner products Mar 9, 2020 at 0:25
• @MaximilianJanisch I see, that's why books are skipping proofs. Why does it hold true though? Just can't get the intuition. Mar 9, 2020 at 0:25
• There is nothing to prove. A positive definite bilinear function is called an inner product iff $\langle x,y\rangle =\overline{\langle y,x\rangle}$ is true. Otherwise, it is not called an inner product Mar 9, 2020 at 0:26
• We can't tell you why it holds true in general. If you specify a particular inner product space then people can help you with this. Are there any examples of inner product spaces that you are familiar with, or will be studying? Mar 9, 2020 at 0:29
• @MaximilianJanisch Not quite bilinear, but rather sesquilinear (since $\langle ax,y\rangle=\overline a\langle x,y\rangle$).
– user239203
Mar 9, 2020 at 1:03

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.

• tt should be noted that there is a divide among mathematicians on what is “the” standard inner product on $\mathbb{C}^n$; your is linear in the second variable and conjugate linear in the first, but many authors define a sesquilinear inner product to be linear in the first coordinate. Mar 9, 2020 at 1:00
• @ArturoMagidin: duly noted. I'm trying to keep it simple here. Cheers! Mar 9, 2020 at 1:02
• @ArturoMagidin: anyway, you saved me a little typing. Thanks again! Mar 9, 2020 at 1:04

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