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My textbook claims that from the axioms for the complex inner product:

$$\left<y,x\right>=\overline{\left<x,y\right>}\tag{1}$$ $$c\left<x,y\right> = \left<cx,y\right>\tag{2}$$

we can derive:

\begin{align} \left<x,cy\right> &= \overline{\left<cy,x\right>}\\ &= \overline{c}\overline{\left<y,x\right>}\\ &= \overline{c}\left<x,y\right> \end{align}

I understand the first and last steps of the derivation, but in the middle step, I don't understand what justifies bringing the $c$ out of the inner product and taking its complex conjugate.

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    $\begingroup$ Should (1) read $\left<x,y\right>=\overline{\left<y,x\right>}$ ? $\endgroup$ – Henry May 22 '17 at 3:06
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$\langle cy, x\rangle = c\langle y,x\rangle$ by the second rule, and so $\overline{\langle cy, x \rangle} = \overline{c\langle y,x \rangle} = \overline{c}\overline{\langle y,x\rangle}$. That last equality just follows from the fact that for any $z,w\in \mathbb C$, we have $\overline{z\cdot w} = \overline{z}\cdot\overline{w}$.

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The definition of an inner product has that its sesquilinear -- linear in one of the arguments, conjugate linear in the other argument. The pulling $c$ out with the conjugate is part of the sesquilinearity.

(The definition also has that its hermitian; if you conjugate the inner product, you reverse the order of taking it)

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    $\begingroup$ Note that there can be some confusion caused by which argument is the one in which the product is conjugate linear and which is linear. This is discussed in the Wikipedia article on inner product spaces for example. $\endgroup$ – Brian Borchers May 22 '17 at 0:14
  • $\begingroup$ Yes, you just need to be aware of the convention in whatever you're reading. $\endgroup$ – Batman May 22 '17 at 0:44

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