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.
Best Answer
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 $$<v,v'>_V=<L(v),L(v')>_W:=<w,w'>_W $$ (##)
If this is the case, we just use the fact that every vector can be expanded using a basis, and that the inner-product is bilinear. Let :
$v=c_1v_1+....+c_nv_n ; v'=c'_1v_1+....+c'_nv_n$ and
$L(v)=w=k_1w_1+.....+k_nw_n ; L(v')=w'=k'_1w_1+....+k'_n v_n$
Write this inside of (##) and use bilinearity of the inner-product, and it should all fall into place.