Dual Vector Spaces and Scalar Products

Remember that any linear map is fully determined by its action on an (arbitrary) basis. In fact, for ~v = ∑ 1≤k≤n λk~vk one gets νi(~v) = λi ∈ R (i = 1, . . . , n). We prove that ν1, . . . , νn ∈ V ∗ are linearly independent. Assume that the vector α := ∑ 1≤k≤n μkνk ∈ V ∗ is the zero map. I.e. ν(~v) = 0 ∈ R holds for all ~v ∈ V . Since this holds for all ~v ∈ V , it follows that μ1 = μ2 = · · ·μn = 0 simply by letting ~v = ~vk, k = 1, . . . , n. Hence, n ≤ dim(V ∗). To demonstrate that span(ν1, . . . , νn) = V ∗, we remember that the vector space HomR(V,W ) of all linear maps between real vector spaces V and W (each of finite dimension) equals dim(V )dim(W ). In fact, choosing a basis in V and a basis in W allows to identify V with R ) and W with R . Hence, HomR(V,W ) can be identified with the real vector space R )×dim(W ) of dimension dim(V )dim(W ). In the special case where W = R one therefore obtains that dim(V ∗) = dim(V ) as was to be proven.