v2009.01.01 - Convex Optimization

658 APPENDIX E. PROJECTION
Because 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 that
yields a particular point x∈ R(A) whenever
R(A) = aff K = R(a 1 ) ⊕ R(a 2 ) ⊕ ... ⊕ R(a n ) (1815)
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 suppose
M = R(u) ⊕ Y (1816)
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 (1817)
nonorthogonal projection of x on R(u) can be determined from
nonorthogonal projection of x on Y , and vice versa.
Such a scenario is realizable were there some arbitrary basis for Y
populating a full-rank skinny-or-square matrix A
A ∆ = [ basis Y u ] ∈ R n+1 (1818)
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 projector
whereas P Y and P u are nonorthogonal projectors.
Now suppose, for example, P Y is an elementary matrix (B.3); in
particular,
P Y = I − e 1 1 T =
[0 √ ]
2V N ∈ R N×N (1819)
where Y = N(1 T ) . We have M = R N , A = [ √ 2V N e 1 ] , and u = e 1 .
Thus P u = e 1 1 T is a nonorthogonal projector projecting on R(u) in a
direction parallel to a vector in Y (E.3.5), and P Y x = x − e 1 1 T x is a
nonorthogonal projection of x on Y in a direction parallel to u .