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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?

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    $\begingroup$ This is an axiom for inner products $\endgroup$ Mar 9, 2020 at 0:25
  • $\begingroup$ @MaximilianJanisch I see, that's why books are skipping proofs. Why does it hold true though? Just can't get the intuition. $\endgroup$
    – shine
    Mar 9, 2020 at 0:25
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    $\begingroup$ 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 $\endgroup$ Mar 9, 2020 at 0:26
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    $\begingroup$ 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? $\endgroup$ Mar 9, 2020 at 0:29
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    $\begingroup$ @MaximilianJanisch Not quite bilinear, but rather sesquilinear (since $\langle ax,y\rangle=\overline a\langle x,y\rangle$). $\endgroup$
    – user239203
    Mar 9, 2020 at 1:03

2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$ Mar 9, 2020 at 1:00
  • $\begingroup$ @ArturoMagidin: duly noted. I'm trying to keep it simple here. Cheers! $\endgroup$ Mar 9, 2020 at 1:02
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    $\begingroup$ @ArturoMagidin: anyway, you saved me a little typing. Thanks again! $\endgroup$ Mar 9, 2020 at 1:04
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Suppose $w, z \in \mathbb{C}^n$. Using the additive properties of complex conjugation, $$ \langle z, w \rangle = \sum_{i = 1}^{n} z_i \overline{w}_i = \overline{\overline{\sum_{i = 1}^{n} z_i \overline{w}_i}} = \overline{\sum_{i = 1}^n \overline{z_i \overline{w}_i}} = \overline{\sum_{i = 1}^{n} \overline{z}_i w_i} = \overline{\sum_{i = 1}^{n} w_i \overline{z}_i} = \overline{\langle w, z \rangle} $$

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