Is there only one inner product?

I've read a book about linear algebra, which does not rush into introducing vectors as a list of numbers, i.e. looking at axioms only. Looking at it this way, it seemed that the dot product $\sum x_iy_i$ is not a definition of the inner product, but rather can be derived from the inner product axioms and is the only inner product (in a sense). Is this true?

If you just take the vector axioms and the inner product axioms, then vectors are just objects and there is no mention of lists of numbers. It is only when you decide on a representation with a basis, that lists of numbers appear. To me it seems, that you can always choose an orthonormal basis which implies $\sum x_iy_i$. Therefore this dot product can be derived from the axioms, it is not an arbitrary definition and it is the only one. Or in other words, if you make up a different equation for the inner product, then your basis is probably not orthonormal and you should have chosen a different basis instead.

Therefore, is $\sum x_iy_i$ the only inner product given that you choose an appropriate basis?

(What changes if you use a different signature for the multiplication, i.e. vectors square to -1 or 0? Is there something additional to consider if you use a complex vector space?)

Solution 1:

Assuming that a vector space $V$ is finite-dimensional, then, yes, if $\langle\cdot,\cdot\rangle$ is an inner product defined on $V$, then there is some basis $\{v_1,v_2,\ldots,v_n\}$ of $V$ (where $n=\dim V$) such that if $v=\sum_{k=1}^n\alpha_kv_k$ and $w=\sum_{k=1}^n\beta_kv^k$, then$$\langle v,w\rangle=\sum_{k=1}^n\alpha_k\beta_k.$$Simply take any basis $\{e_1,e_2,\ldots,e_n\}$ of $V$, apply the Gram-Schmidt orthogonalization to it, and then you will the basis $\{v_1,v_2,\ldots,v_n\}$ that you're after.

Solution 2:

To add to the answer that José Carlos Santos gave, if you have two inner products $\langle\cdot\,,\cdot\rangle_A$ and $\langle\cdot\,,\cdot\rangle_B,$ then you can have bases $E=\{e_0,e_1,...,e_n\}$ such that $F=\{f_0,f_1,...,f_n\}$ such that $E$ is orthonormal with respect to $\langle\cdot\,,\cdot\rangle_A$ and $F$ is orthonormal with respect to $\langle\cdot\,,\cdot\rangle_B,$ yet $E\neq{F},$ rendering $\langle\cdot\,,\cdot\rangle_A\neq\langle\cdot\,,\cdot\rangle_B.$ So, no, there is not "only one" inner product on any given vector space with a base. The inner products could be isomorphic, but need not be. As for how many inner products a vector space can have, it can have infinitely many, at least if the vector space has finite dimension.