(Orthogonal Projectors) This exercise generalizes the idea expressed in Exercise 2.5. An orthogonal projector can be thought of as simply cancelling the contributions to a vector of some elements of an orthonormal basis. That is, if {b₁,..., b_N} is an orthonormal basis of R^N (so that ||b_n|| = 1 and b_n^'b_m = 0 for all n, m = 1,..., N, n ≠ m) the orthogonal projection of a vector z = Σ_n=1^N α_nb_n

onto the subspace spanned by {b₁,..., b_M}, M < N, is simply Σ_n=1^M α_nb_n.

Let P be an orthogonal projector onto a K-dimensional subspace of R^N.

(a) Given only P, how can a matrix B₁ be found that i. is full-column rank, ii. B₁^'B₁ = I_K, and iii. P = B₁B₁^'?

(Hint: Exercise 2.13.)

(b) Show how to find a second matrix B₂ that i. is full-column rank, ii. B₂^'B₂ = I_N-K, iii. B₂^'B₁ is an (N − K) × K matrix of zeros, and iv. B = [B₁, B₂] is nonsingular.

(Hint: Consider the orthogonal projector I − P.)

(c) Show that for every z ∈ R^N, z = Ba = B₁α₁ + B₂α₂ for some α ∈ R^N.

(d) Show that Pz = B₁α₁, confirming the interpretation of orthogonal projectors given above.