condition 4315 Let 1 2 be inner products on a finitedimensio

condition 4.3.15:

Let (,)_1, (,)_2 be inner products on a finite-dimensional real vector space V. Use Exercise 4.3.15 and the spectral theorem to show that there exists a basis of V, which is orthogonal (but perhaps not orthonormal) with respect to both inner products simultaneously. Let alpha = {x_1, ..., x_n} be a basis in R^n, and let B be a bilinear mapping. Define [B]_alpha to be the n times n matrix [B]_alpha = (B(x_t, x_j)). That is, the (i, j) entry of [B]_alpha is the scalar B(x_l, x_j). Show that for all x, y element of R^n B(x, y) = [x]^t_alpha middot [B]_alpha middot [y]_alpha (This shows that every bilinear mapping on R^n is of the kind we saw in Exercise 14.) If B is the standard inner product, and alpha is the standard basis in R^n, what is [B]_alpha?

Solution

Apply the Theorem to the two bilinear forms given by

               <x,y>₁ ,<x,y>₂

(By definition an inner product is a bilinear form)

Let B₁ and B₂ be the matrices obtained by 4.3.15

Then B₁ and B₂ are commuting symmetric matrices and hence they can be simultaneously diagonalized.

In other words there exists a common basis of eigen vectors {v[1],....,v[n]} for B₁ and B₂,

As eigenvectors for distinct eigevalues are orthogonal , it follows that the eigenvectors are orthogonal wrt both inner products.