Since an inner product space is an abstract vector space with an additional structure called an inner product, and this additional structure is a component wise operation that associates each pair of vectors in the space to a scalar, then:

is it right to say that this operation can be different depending of the kind of vector space ($\mathbb{R}^{n}$, Euclidean spaces, functional spaces)?

is it right to say that I can define this operation following my personal defined rule as long as the resulting inner vector space satisfy the rules of linearity and other inner product general properties?

Since from an abstract algebra point of view a vector of the vector space is a set V over a field F in which V is an Abelian group $G_{i}$ under addition with certain properties such that for each $\alpha\in F$ and $\boldsymbol{v}_{i}\in V$ also $\alpha\boldsymbol{v}_{i}\in V$ then:

is it right to say that in a general form the inner vector operation between two vectors of the space $\left \langle \alpha \boldsymbol{v}_{1}, \beta \boldsymbol{v}_{2} \right \rangle$ gives a result of the kind $\alpha \beta \left \langle \boldsymbol{v}_{1} ,\boldsymbol{v}_{2} \right \rangle$ and this involves that the my personal defined rule applies only to $\left \langle \boldsymbol{v}_{1} ,\boldsymbol{v}_{2} \right \rangle$?

  • 1
    $\begingroup$ 1) Yes, definitely. 2) Well, yes. You can define $\langle v_1,v_2\rangle$ in any way you like, as long as the resulting operation satisfies the axioms of inner product. In particular, any such definition must satisfy the mentioned identity. $\endgroup$ – Marcin Łoś Jul 23 '14 at 14:00
  • $\begingroup$ The inner product need not be defined "componentwise". Consider for instance the space of continuous functions on $[0,1]$ with inner product given by the integral of the product. $\endgroup$ – lhf Jul 23 '14 at 14:10

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