Question

T preserves inner product t is an isomorphism

Answer 1

10down votefavorite5I think I understand why all finite-dimensional vector spaces over a field KK are isomorphic to KnKn. Any linear map T:V→WT:V→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 RnRn with standard basis {ei}ni=1{ei}i=1n and the natural Euclidean inner-product⟨x¯,y¯⟩=∑i=1nxiyi⟨x¯,y¯⟩=∑i=1nxiyiOn the other hand, there's Pn([−1,1])Pn([−1,1]), the space of real polynomials on [−1,1][−1,1] of degree less than nn, with the obvious basis {1,x,x2,...,xn−1}{1,x,x2,...,xn−1} and the L2L2 inner-product⟨p,q⟩=∫−1−1p(x)q(x)dx⟨p,q⟩=∫−1−1p(x)q(x)dxThey'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 L2L2 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 PnPn corresponding to the standard basis on RnRn. 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 TT 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 L2L2 inner-product on Pn([−1,1])Pn([−1,1]), I have to map {ei}ni=1{ei}i=1n to the Legendre polynomials. If I want the monomial basis {1,x,x2,...,xn−1}{1,x,x2,...,xn−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 PnPn 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 VV, and I map that to a basis for WW. Is there an inner-product on WW such that this latter basis is orthonormal in WW? More generally, is there an inner-product on WW that acts the same on WW as the inner-product on VV acts on VV?I have a feeling that I'm confused about some pretty fundamental things here.linear-algebra hilbert-spaces inner-product-space orthogonal-polynomials orthonormalshareciteimprove this questionedited Apr 5 '13 at 19:02asked Apr 5 '13 at 7:08AndrewG1,515828   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 nn, and in particular the monomial xnxn, out from Pn([−1,1])Pn([−1,1]) in the question, since it appeared to me that you intended to define a space of dimension nn. – Marc van LeeuwenApr 5 '13 at 8:21   @Marc: Woops, of course. Thanks. – AndrewG Apr 5 '13 at 19:02add a comment2 Answersactiveoldestvotesup vote5down voteIf two vector spaces VV and WW are isomorphic, there are as many isomorphisms V→WV→W as automorphisms of WW. Indeed, is Iso(V,W)Iso(V,W) is the set of isomorphisms V→WV→W, Aut(W)Aut(W) is the set of all automorphisms of WW and f0∈Iso(V,W)f0∈Iso(V,W) is any isomorphism, the functiona∈Aut(W)↦a∘f0∈Iso(V,W)a∈Aut(W)↦a∘f0∈Iso(V,W)is a bijection. It follows there are as many isomorphisms from VV to WW as there are automorphisms of WW. A similar argument shows that there are as many such isomorphisms as there are automorphisms of VV, too.Let ⟨−,−⟩0⟨−,−⟩0 be an inner product on a vector space VV., and let as before Aut(V)Aut(V) be the set of all automorphisms of VV and let Inn(V)Inn(V) be the set of all inner products on VV. Then for every f∈Aut(V)f∈Aut(V) there is an inner product ⟨−,−⟩f⟨−,−⟩f on VV such that for all v,w∈Vv,w∈V we have⟨v,w⟩f=⟨f(u),f(w)⟩0,⟨v,w⟩f=⟨f(u),f(w)⟩0,and the functionf∈Aut(V)↦⟨−,−⟩f∈Inn(V)f∈Aut(V)↦⟨−,−⟩f∈Inn(V)is surjective; in this way we obtain a description of all inner products on VV. It is not injective, though.Indeed, two automorphisms ff, g∈Aut(V)g∈Aut(V) have the same image, so that ⟨f(v),f(w)⟩0=⟨g(v),f(w)⟩0⟨f(v),f(w)⟩0=⟨g(v),f(w)⟩0 for all vv, w∈Vw∈V if and only if the composition f∘g−1f∘g−1preserves the original inner product ⟨−,−⟩0⟨−,−⟩0, in the sense that⟨(f∘g−1)(v),(f∘g−1)(w)⟩0=⟨v,w⟩0⟨(f∘g−1)(v),(f∘g−1)(w)⟩0=⟨v,w⟩0for all vv, w∈Vw∈V.