Isomorphisms of inner-product spaces I think I understand why all finite-dimensional vector spaces over a field $\mathbb{K}$ are isomorphic to $\mathbb{K}^n$. Any linear map $T: V \rightarrow W$ between finite-dimensional vector spaces taking a basis to a basis is automatically an isomorphism, by linearity. (c.f. this nice post.)
But I'm puzzled about the following.
On the one hand, there's $\mathbb{R}^n$ with standard basis $\{e_i\}_{i=1}^n$ and the natural Euclidean inner-product $$\langle \bar{x},\bar{y}\rangle = \sum_{i=1}^n x_i y_i$$
On the other hand, there's $P_n([-1,1])$, the space of real polynomials on $[-1,1]$ of degree less than $n$, with the obvious basis $\{1,x,x^2,...,x^{n-1}\}$ and the $L^2$ inner-product $$\langle p,q\rangle = \int_{-1}^{-1} p(x)q(x)dx$$
They're both Hilbert spaces. The basis given for the former is orthonormal; the latter is not (but we can apply Gram-Schmidt to build the Legendre polynomials, which are.)
This seems somehow strange to me: the $L^2$ inner-product looks like the most straightforward generalization of the Euclidean inner-product to function spaces, and the basis of monomials seems like the most natural basis of $P_n$ corresponding to the standard basis on $\mathbb{R}^n$. The Legendre polynomials, by contrast, appear bizarre and complicated. The vector spaces are obviously isomorphic: given any basis of each, we can easily construct an isomorphism $T$ mapping each basis to the other. But in the above example, orthonormality isn't preserved.
If I want to keep orthonormality, it seems I have to choose: if I want the $L^2$ inner-product on $P_n([-1,1])$, I have to map $\{e_i\}_{i=1}^n$ to the Legendre polynomials. If I want the monomial basis $\{1,x,x^2,...,x^{n-1}\}$, I have to pick a different inner-product. I can't have my cake and eat it, too. (And I don't even know if an inner-product on $P_n$ exists for which the basis of monomials is orthonormal.)
This leads me to several questions.

  
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*For isomorphic, finite-dimensional vector spaces V and W, just how many isomorphisms are there?
  
*How many distinct inner-products can there be?
  
*Is there some sort of 'natural' correspondence here between isomorphisms and pairs of inner-products? (This was just confusion on my part.)
  
*Suppose I specify an inner-product and an orthonormal basis for $V$, and I map that to a basis for $W$. Is there an inner-product on $W$ such that this latter basis is orthonormal in $W$? More generally, is there an inner-product on $W$ that acts the same on $W$ as the inner-product on $V$ acts on $V$?
  

I have a feeling that I'm confused about some pretty fundamental things here.
 A: Just a few fairly simple remarks should settle most questions.


*

*A finite dimensional inner product space always has at least one orthonormal basis.

*A linear map $V\to W$ between two finite dimensional inner product spaces is an isomorphism of inner product spaces if and only if the image of a (some fixed) orthonomal basis of $V$ is an orthonormal basis of $W$.

*If $V$ is a real vector space of dimension $n$, and $\def\B{\mathcal B}\B$ a basis of $V$, then there is a unique inner product on $V$ for which this basis is orthonormal. (This answers your last question) With $f_\B:V\to \Bbb R^n$ the map sending a vector $v$ to its coordinates in $\mathcal B$, this inner product is obtained from the  by transport of structure via $f_\B$, in other words $\langle v,w\rangle_V=\langle f_\B(v),f_\B(w)\rangle_{\Bbb R^n}$ by definition. (By the previous point applied to $f_\B^{-1}$ the basis $\mathcal B$ is orthonormal if and only if $f_\B$ is an isomorhism of inner product spaces, and the equation used as definition states just that.)


So up to isomorphism $\Bbb R^n$ with the standard inner product is the unique $n$-dimensional inner product space. For two inner product spaces $V,W$ of dimension $n$, there are as many isomorphisms of inner product spaces as there are orthonomal bases in $W$ (or in $V$); the set is also in bijection with the set of automorphsims of the standard inner product space $\Bbb R^n$, which is the orthogonal group $O_n(\Bbb R)$. On a given space $V$ there are as many different (though isomorphic) inner products as there are cosets in $O_n(\Bbb R)\backslash GL_n(\Bbb R)$.
(Both the sets of isomorphisms and of different inner products are infinite in general, so "as many" should be interpreted as "naturally in bijection with".)
Explanation for the latter correspondence: fixing $\B$, the cooordinate map $V\to \Bbb R^n$ corresponding to any basis of $V$ is of the form $g\circ f_\B$ for some $g\in GL_n(\Bbb R)$, which defines an inner product on $V$, and $g_1,g_2\in GL_n(\Bbb R)$ define the same inner product iff the standard basis of $\Bbb R^n$ transported to $V$ by $(g_1\circ f_\B)^{-1}$ and then back to $\Bbb R^n$ by $g_1\circ f_\B$ give an orthonormal basis, which means $g_2\circ g_1^{-1}\in O_n(\Bbb R)$ of $g_2\in O_n(\Bbb R)g_1$. The set of inner products on $\Bbb R^n$ is also in bijection with the set of positive definite symmetric $n\times n$ matrices.
A: *

*If two vector spaces $V$ and $W$ are isomorphic, there are as many isomorphisms $V\to W$ as automorphisms of $W$. Indeed, is $Iso(V,W)$ is the set of isomorphisms $V\to W$, $Aut(W)$ is the set of all automorphisms of $W$ and $f_0\in Iso(V,W)$ is any isomorphism, the function $$a\in Aut(W)\mapsto a\circ f_0\in Iso(V,W)$$ is a bijection. It follows there are as many isomorphisms from $V$ to $W$ as there are automorphisms of $W$. A similar argument shows that there are as many such isomorphisms as there are automorphisms of $V$, too.


*Let $\langle\mathord-,\mathord-\rangle_0$ be an inner product on a vector space $V$., and let as before $Aut(V)$ be the set of all automorphisms of $V$ and let $Inn(V)$ be the set of all inner products on $V$. Then for every $f\in Aut(V)$ there is an inner product $\langle\mathord-,\mathord-\rangle_f$ on $V$ such that for all $v,w\in V$ we have $$\langle v,w\rangle_f=\langle f(u),f(w)\rangle_0,$$ and the function $$f\in Aut(V)\mapsto \langle\mathord-,\mathord-\rangle_f\in Inn(V)$$ is surjective; in this way we obtain a description of all inner products on $V$. It is not injective, though.
Indeed, two automorphisms $f$, $g\in Aut(V)$ have the same image, so that $\langle f(v),f(w)\rangle_0=\langle g(v),f(w)\rangle_0$ for all $v$, $w\in V$ if and only if the composition $f\circ g^{-1}$ preserves the original inner product $\langle\mathord-,\mathord-\rangle_0$, in the sense that $$\langle (f\circ g^{-1})(v),(f\circ g^{-1})(w)\rangle_0=\langle v,w\rangle_0$$ for all $v$, $w\in V$.


*Suppose $V$ is a vector space with an inner product $\langle\mathord-,\mathord-\rangle$ and that $W$ is a vector space. Suppose, moreover, that $f:W\to V$ is an isomorphism of vector spaces. Then we can define a new inner product $\langle\mathord-,\mathord-\rangle'$, now on $W$, so that for all $v$, $w\in W$ we have $$\langle v,w\rangle'=\langle f(v),f(w)\rangle.$$ With respect to the inner products on $V$ and on $W$ that we now have, the map $f$ is an isomorphism of inner product spaces.
This answers your 4th question.


*I don't understand your 3rd question :-)
