I have a question: Does every vector space have an inner product?
I think, yes. But I failed to find an essential reason. If it does not exist, then give me a counterexample. Thanks.
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If a space is an inner product space it satisfies Parallelogram law. Can you find an example of a normed vector space where this law doesn't hold? Hint: $L^p$. – Robert Cardona Oct 16 '14 at 15:23
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I saw this post. But I don't understand... So, I want an easier example than that. I cannot have tried because this question is so abstract... we finally assume that the given vector space has the finite dimension. – user184859 Oct 16 '14 at 15:24
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Take a vector space $X$ over $\mathbb R$ (or $\mathbb C$) and fix a basis $\{v_i\}_{i\in I}$.
If $x,y\in X$, then they can be uniquely expressed as $$ x=\sum_{i\in I}c_iv_i,\quad y=\sum_{i\in I}d_iv_i, $$ with each of the sum above finite.
Define $$\langle u,v\rangle=\sum_{i\in I}c_id_i. $$ This is an inner product in $X$.
In the case of $\mathbb C$, the definition is $$\langle u,v\rangle=\sum_{i\in I}c_i\overline{d_i}. $$
Belgi
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Yiorgos S. Smyrlis
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Yeah, I see. Should I define the inner product over real or complex? If not, is it possible that an inner product does not exist? – user184859 Oct 16 '14 at 15:31
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Inner product is always definable. You should choose real or complex depending on the field over which your vector space is defined on. – Yiorgos S. Smyrlis Oct 16 '14 at 15:37
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Inner products are defined for spaces over $\mathbb{R}$ or over $\mathbb{C}$ only.
The reason is that in the definition we require $$\langle u,v \rangle=\overline{\langle v,u\rangle}$$
and what is the meaning of $$\overline{\langle v,u\rangle}$$
when $\langle v,u \rangle$ is not complex ?
Belgi
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If the given scalar is neither real nor complex, then some inner product may exists in the different way? If only the scalar is complex(real), the above definition, I think, is working. – user184859 Oct 16 '14 at 15:27