A question about inner products on abstract vector spaces I have been reading some materials and, for the n-th time in my life, there was a definition of an inner product as a function $V \times V \rightarrow F$, where $V$ is an abstract vector space and $F$ is an underlying scalar field.
However, it got me thinking. Inner product is this special function which gives us some number, right? In order to get a number, you must work with numbers.
Now, in general, our abstract vectors are not sequences of numbers, be it matrices, ordered pairs, polynomials or whatever, they are just that, abstract vectors. The only thing that connects these vectors with some numbers are their coordinates with respect to some basis. And in that case, we performed an isomorphism to the vector space $R^n$ (if n is the dimension of the vector space), so we effectively consider only $R^n$ in that case, so let's not do that.
What I am getting here, is how is it possible to construct an inner product on an abstract vector space? How do we take two ordinary abstract vectors and get a number out of it? You can't sum these vectors, you can't multiply them etc., they are not numbers. You can only do that with their coordinates. And if you do that, then you are not defining an inner product on a general vector space, you have defined an inner product on $R^n$ and you are using isomorphism to indirectly define inner product on other vector spaces of the same dimension. That can't be right, can it?
 A: You can pick a basis and define the inner product by specifying what it does to a basis. 
But this is somewhat unsatisfactory. In practice, how you go about writing down a meaningful inner product on $V$ depends on how $V$ itself is constructed. For example, if $V$ is, say, a space of real-valued functions on some measure space $(X, \mu)$, then a natural inner product to write down is
$$\langle f, g \rangle = \int_X f(x) g(x) \, d \mu$$
provided that this integral always converges. 
A: Let $V$ be an arbitrary linear space over $F\in\{\mathbb{R},\mathbb{C}\}$. Consider some Hamel basis of $V$, denote it $\{e_\lambda:\lambda\in\Lambda\}$. Then for each $v\in V$ we have a family of numbers $\{v_\lambda:\lambda\in\Lambda\}\subset F$ such that
$$
v=\sum\limits_{\lambda\in\Lambda} v_\lambda e_\lambda\tag{1}
$$
Note that only finitely many numbers in $\{v_\lambda:\lambda\in\Lambda\}$ are non-zero, so $(1)$ is well defined. Then you can define inner product by
$$
\langle v,w\rangle=\begin{cases}\sum\limits_{\lambda\in\Lambda} v_\lambda w_\lambda &\quad\text{ if }\quad F=\mathbb{R}\\\sum\limits_{\lambda\in\Lambda} v_\lambda \overline{w_\lambda} &\quad\text{ if }\quad F=\mathbb{C}
\end{cases}\tag{2}
$$
One can check that $\langle\cdot,\cdot \rangle :V\times V\to F$ is well defined inner product.
