Is the standard scalar product in a coordinate space basis independent? Would you say that the standard scalar product in $K^n$, $\left< x,y \right>=\sum_i x_i y_i$, is basis-independent or not ?
I would argue that it is, because we don't use the components of the vectors $x,y,$ to define this function, but the vectors themselves (although there is a basis, namely the standart basis, which, if we were to define the standard scaler product from the beginning in a basis-dependent manner - i.e. as $\left< x,y \right>_ {\mathcal{B}}=\sum_i ([x]_{\mathcal{B}})_i ([y]_{\mathcal{B}})_i$, where $\mathcal{B}$ denotes some basis of $K^n$ and $([x]_{\mathcal{B}})_i$ the $i$-th component of the coordinate vector of $x$ w.r.t. the basis $\mathcal{B}$ - would give the same function).
 A: In a sense everything in $K^n$ is basis independent, because vector spaces of the very special kind $K^n$ don't actually need any basis in order to associate coordinates to vectors: vectors in $K^n$ are $n$-tuples of numbers, which components one can access directly. Yes, there is  the standard basis, but it is given rather than chosen; moreover this is really a device similar to inventing, in physics, a unit (namely $1$) for dimensionless quantities that don't need a unit: the coordinates of a vector in the standard basis are just the components of the vector itself, just like a dimensionless quantity $x$ expressed in the unit$~1$ just gives back the number$~x$. The standard basis itself is always the same, so "basis independent". This is the sense in which  in $K^n$ basis independence means nothing. These spaces are the worst possible examples to learn about the meaning of bases, just like dimensionless quantities are bad examples to learn about the meaning of physical units. (Unfortunately, they are also the easiest examples to talk about concretely.)
However, in any other kind of vector space $V$, basis independence means that expressing things in coordinates, after choosing a basis to get an isomorphism $V\to K^n$, gives the same formulas regardless of which basis was used. Now very few simple things are basis independent, apart from the zero vector. However some more complicated things which use the same basis more than once, are basis independent because the various effects of the basis choice cancel out. The characteristic polynomial of linear operator (or its coefficients, like determinant and trace) is an example of a basis independent quantity.
The standard inner product applied to the coordinates of two vectors in $V$ expressed in a basis does involve that basis more than once, but the effects do not cancel out in general. For instance if one scales the basis by a factor $f$, then the inner products so defined get to be divided by$~f^2$. Therefore the inner product on $V$ that one gets does depend on the choice of basis, and there is no such thing as a standard inner product on a vector space not equipped with additional structure. I should note that the effect of changing basis does cancel out in some cases: when changing to a basis that was orthonormal for the initial inner product, then the new inner product will coincide with the old one. There are quite a few orthonormal bases, but of course there are even more bases that are not orthonormal.
A: Unlike vector spaces arising in other ways, ${\mathbb R}^n$ has a preferred basis, namely the standard basis. Declaring this basis as orthonormal defines the standard scalar product
$$\langle x,y\rangle:=\sum_{k=1}^n x_k y_k\tag{1}$$
on ${\mathbb R}^n$. This scalar product is a well defined function on ${\mathbb R}^n\times{\mathbb R}^n$ which does not change its values under a change of basis. What is base dependent, however, is the way that this scalar product is computed. When you replace the standard basis $(e_i)_{1\leq i\leq n}$ by some other basis $(f_k)_{1\leq k\leq n}$ the points $x$ of our space obtain new coordinates $(\bar x_1,\ldots ,\bar x_n)$. In terms of these new coordinates the scalar product is no longer computed by a simple formula of type $(1)$, but by means of a more complicated formula
$$\langle x,y\rangle:=\sum_{i, \ j=1}^n g_{ij}\bar x_i\bar y_j\ ,$$
whereby the constant matrix $G:=[g_{ij}]$ depends in a characteristic way on the transformation matrix between the two bases.
