v2007.09.13 - Convex Optimization

E.5. PROJECTION EXAMPLES 597of relative dual cone K ∗ ∩aff K=cone(A †T ) (2.13.9.4) correspond to thelinearly independent (B.1.1.1) rows of A † . Directions of nonorthogonalprojection are determined by the pseudoinverse; id est, direction ofprojection a i a ∗Ti x−x on R(a i ) is orthogonal to a ∗ i . E.8Because the extreme directions of this cone K are linearly independent,the component projections are unique in the sense:there is only one linear combination of extreme directions of K thatyields a particular point x∈ R(A) wheneverR(A) = aff K = R(a 1 ) ⊕ R(a 2 ) ⊕ ... ⊕ R(a n ) (1702)E.5.0.0.4 Example. Nonorthogonal projection on elementary matrix.Suppose P Y is a linear nonorthogonal projector projecting on subspace Y ,and suppose the range of a vector u is linearly independent of Y ; id est,for some other subspace M containing Y supposeM = R(u) ⊕ Y (1703)Assuming P M x = P u x + P Y x holds, then it follows for vector x∈MP u x = x − P Y x , P Y x = x − P u x (1704)nonorthogonal projection of x on R(u) can be determined fromnonorthogonal projection of x on Y , and vice versa.Such a scenario is realizable were there some arbitrary basis for Ypopulating a full-rank skinny-or-square matrix AA ∆ = [ basis Y u ] ∈ R n+1 (1705)Then P M =AA † fulfills the requirements, with P u =A(:,n + 1)A † (n + 1,:)and P Y =A(:,1 : n)A † (1 : n,:). Observe, P M is an orthogonal projectorwhereas P Y and P u are nonorthogonal projectors.Now suppose, for example, P Y is an elementary matrix (B.3); inparticular,P Y = I − e 1 1 T =[0 √ ]2V N ∈ R N×N (1706)E.8 This remains true in high dimension although only a little more difficult to visualizein R 3 ; confer , Figure 50.