v2004.06.19 - Convex Optimization

E.5. PROJECTION EXAMPLES 399where each dyad a i a † i is a nonorthogonal projector on R(a i ) . (§E.6.1) Theextreme directions of K=cone(A) are {a 1 ,..., a n } the linearly independentcolumns of A while {a †T1 ,..., a †Tn } the extreme directions of relative dualcone K ∗ ∩aff K= cone(A †T ) (§2.9.1) correspond to the linearly independent(§B.1.1.1) rows of A † . The directions of nonorthogonal projection are determinedby the pseudoinverse; id est, direction of projection a i a † i x−x onR(a i ) is orthogonal to a †Ti . 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 ) (1233)E.5.0.1.4 Example. Nonorthogonal projection on elementary matrix.Suppose P Y is a linear nonorthogonal projector on subspace Y ⊂ M , andsuppose the range of a vector u is linearly independent of Y ; id est, forsome other subspace M ,M = R(u) ⊕ Y (1234)Assuming P M x = P u x + P Y x holds, then it follows for vector x∈M ,P u x = x − P Y x , P Y x = x − P u x (1235)the nonorthogonal projection of x on R(u) can be determined from thenonorthogonal projection of x on Y , and vice versa.Such a scenario is realizable were there some arbitrary basis for Y populatinga full-rank skinny-or-square matrix A ,A ∆ = [ basis Y u ] ∈ R n+1 (1236)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.E.8 This remains true in high dimension although only a little more difficult to visualizein R 3 ; confer , Figure 2.24.