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Let $X$ be the set of all $n$-tuples of complex numbers and let $x,y \in X$ such that $x=(a_1,a_2,\dots,a_n)$ and $y=(b_1,b_2,\dots,b_n)$. The inner product of the two vectors is given by $\langle x,y\rangle=\sum_i a_i \bar b_i$. To show that this is an inner product space(IPS), it must satisfy three axioms. My question is this:

Assuming that the vector space is real, that is, each element of $X$ is an $n$-tuple of real numbers, how can I prove the first axiom of IPS: $\langle x,y\rangle=\overline{\langle y,x\rangle}$?

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  • $\begingroup$ I think I have figured out what your question means. Can you please check it over to make sure it is what you intended? $\endgroup$
    – dfeuer
    Oct 2, 2013 at 0:45
  • $\begingroup$ Looking at the original post, I cannot but wonder: did you scan this somewhere and then read it with some OCR software?! $\endgroup$
    – Vedran Šego
    Oct 2, 2013 at 1:08
  • $\begingroup$ Yes, that's what I mean @dfeuer $\endgroup$
    – user73535
    Oct 2, 2013 at 10:04

1 Answer 1

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You don't need the assumption that $X$ is a real space.

$$\langle x, y \rangle = \sum_i a_i \overline{b_i} = \sum_i \overline{\overline{a_i} b_i} = \overline{\sum_i \overline{a_i} b_i} = \overline{\sum_i b_i \overline{a_i}} = \overline{\langle y, x \rangle}.$$

If everything is real, you can freely remove conjugation wherever you want, since $\xi = \overline{\xi}$ for all $\xi \in \mathbb{R}$ (and everything in the above expression is, in that case, real).

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  • $\begingroup$ Ok. Thanks @Vedan Sego $\endgroup$
    – user73535
    Oct 2, 2013 at 13:33

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