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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.

  • 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.

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LaTeX tips: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use \langle and \rangle. Additionally, to get braces inside LaTeX, use \{ and \} (alternatively, \lbrace and \rbrace). –  Zev Chonoles Apr 5 '13 at 7:10
    
Thanks. That was really bothering me, but I wasn't sure how to do it right. –  AndrewG Apr 5 '13 at 7:12
    
I've removed the polynomials of degree $n$, and in particular the monomial $x^n$, out from $P_n([-1,1])$ in the question, since it appeared to me that you intended to define a space of dimension $n$. –  Marc van Leeuwen Apr 5 '13 at 8:21
    
@Marc: Woops, of course. Thanks. –  AndrewG Apr 5 '13 at 19:02
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2 Answers

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.

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I am studying this post carefully because the issue of the "number of "different" field extensions of degree n" has come up in my study group. Are they all isomorphic as a corollary to the discussion above? That seems like an awfully scary conclusion to make off hand. –  David Dyer Oct 22 '13 at 12:47
    
@DavidDyer: The answer is about inner product spaces, which are real vector spaces equipped with an inner product. This is rather unrelated to field extensions. In fact there are no field extensions of $\Bbb R$ of degree $n$, unless $n\in\{1,2\}$ (which goes to show how pathological the field of real numbers is;-). –  Marc van Leeuwen Oct 22 '13 at 12:56
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  1. 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.

  2. 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 $v$, $w\in V$.

  3. 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.

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

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This all looks very helpful. Thank you. I'm going to read, think and sleep on it. (And I think you actually answered my 3rd and 4th question at the same time.) –  AndrewG Apr 5 '13 at 7:33
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