# Isomorphism of inner product spaces

I want to use this in a proof, however I don't know how to prove it itself. I feel as though it's easy to prove by definition but I'm not quite sure..

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

• Which part of the proof are you struggling with? Commented Jun 4, 2014 at 1:45
• It is false: even in $\;\Bbb R^2\;$ with the usual inner product you can construct easily isomorphisms which doesn't respect orthonormality...though an isomorphism must respect the fact that a basis is mapped to a basis. Commented Jun 4, 2014 at 1:45
• Hmmm...but it may be, @Smithson, that this is merely definition: we want isomorphism of inner product vector spaces, so it maybe that it is required $\;\langle Tu,Tv\rangle=\langle u,v\rangle\iff T\;$ is an orthogonal matrix iff it maps orthonormal basis to orthonormal basis... Commented Jun 4, 2014 at 1:48
• But @DonAntonio what my proposition is suggesting is not implicitly statingt that the isomorphisms themselves have to respect orthonormality.. Commented Jun 4, 2014 at 1:52
• Exactly, @Smithson. Then I think it'd be a good idea if you add what your definition of "isomorphism between inner product spaces" is. Commented Jun 4, 2014 at 1:54

For a formal proof, if I understood correctly, you want to show that if a linear map that sends orthonormal bases to orthonormal basis is an inner-product isomorphism, i.e., given $(V;<,>_V,W ; <,>_W$ with respective ortho. bases {$v_i;i=1,..,n$} and {$w_i ; j=1,..,n$} (isomorphic spaces must have the same dimension) with $L(v_i)=w_i$ (we can rename/renumber the basis vectors) , then $$<v,v'>_V=<L(v),L(v')>_W:=<w,w'>_W$$ (##)
$v=c_1v_1+....+c_nv_n ; v'=c'_1v_1+....+c'_nv_n$ and
$L(v)=w=k_1w_1+.....+k_nw_n ; L(v')=w'=k'_1w_1+....+k'_n v_n$