Let V be a finite-dimensional real inner product space and let V^∗ be its dual. Using Theorem 7.10, prove that the map J: V^∗ → V that takes the linear function ℓ ∈ V^∗ to the vector J[ℓ] = a ∈ V satisfying ℓ[v] = (a , v) defines a linear isomorphism between the inner product space and its dual: V^∗ ≃ V.

Transcribed Image Text:

Theorem 7.10. Let V be a finite-dimensional real inner product space. Then every linear function : V → R is given by taking the inner product with a fixed vector a € V: l[v] = (a, v). (7.12) Proof: Let V₁, Vn be a basis of V. If we write v = y₁V₁++ynVn then, by linearity, l[v] = y₁ l[v₁] + +yn [vn] =b₁y₁ + +by where b = [u]. (7.13) On the other hand, if we write a = (a,v) = n i,j=1 ₁ V₁ + *** +In Vn, then 'nn' 72 ₁₁ (V₁, Vj) = Σ gijjis ij=1 (7.14) where G = (9₁5) is the nxn Gram matrix with entries gij = (v₁, v.). Equality of (7.13, 14) requires that Gx = b, where x = (₁, 2,...,xn), b = (b₁,b₂,..., b). Invertibility of G as guaranteed by Theorem 3.34, allows us to solve for x = G-¹b and thereby construct the desired vector a. In particular, if v₁,..., V, is an orthonormal basis, then G = I and hence a = b₁ V₁ + + b₂ Vn