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The conjugate symmetric property of an inner product states that $\langle{x, y}\rangle = \overline{\langle{y, x}\rangle}$. My question is regarding showing this when we don't necessarily know that our field is $\mathbb{C}$.

For a vector space $V$ defined over a field $\mathbb{F}$, let $\langle{,}\rangle: V \times V \rightarrow \mathbb{F}$ be the inner product defined by $\langle{u, v}\rangle = u^t\cdot{}v$.

Then $\langle{u, v}\rangle = u^t\cdot{v} = v^t\cdot{}u = \langle{v, u}\rangle$.

If the field was $\mathbb{R}$, this would be fine. But I'm not sure about the case that the field is $\mathbb{C}$.

I'm not sure how to just throw the conjugate in there. I'm not sure how to conclude that $\langle{u, v}\rangle = \overline{\langle{v, u}\rangle}$.

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    $\begingroup$ What definition are you using for $u^t$? In a general vector space there is no canonical way to define transposes. $\endgroup$ – Brian Fitzpatrick Aug 16 '14 at 20:42
  • $\begingroup$ You can't show that even if the underlying field is $\mathbb C$. $(a,b)=\sum_{i=1}^na_ib_i$ gives an inner product on $\mathbb C^n$ that satisfies $(a,b)=(b,a)$ without conjugation. If you are worried that this does not have all the good properties, how could these properties be there over an arbitrary field anyway? $\endgroup$ – Joonas Ilmavirta Aug 16 '14 at 20:55
  • $\begingroup$ If the field is not $\mathbb{C}$ what is the meaning of conjugation? $\endgroup$ – Conifold Aug 16 '14 at 21:20
  • $\begingroup$ Brian - Well I'm working out of Michael Artin's "Algebra" book, so I was using his definition of going from a row vector to a column vector. Joonas - Are you saying that nothing else remains to be shown? Conifold - I suppose conjugation has no other meaning unless we are working with $\mathbb{C}$. $\endgroup$ – Joseph DiNatale Aug 16 '14 at 22:03
  • $\begingroup$ My point is that if you want to show that property, you need to make some assumptions. What does conjugation mean? What do you assume of the function $\langle\cdot,\cdot\rangle$? Are you interested in all fields? You can't show anything starting from just what you wrote. $\endgroup$ – Joonas Ilmavirta Aug 17 '14 at 10:26

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