v2009.01.01 - Convex Optimization

672 APPENDIX E. PROJECTION
geometric center mapping V(X) = −V XV 1 consistently with (901), then
2
N(V)= R(I − V) on domain S n analogously to vector projectors (E.2);
id est,
N(V) = S n⊥
c (1877)
a subspace of S n whose dimension is dim S n⊥
c = n in isomorphic R n(n+1)/2 .
Intuitively, operator V is an orthogonal projector; any argument
duplicitously in its range is a fixed point. So, this symmetric operator’s
nullspace must be orthogonal to its range.
Now compare the subspace of symmetric matrices having all zeros in the
first row and column
S n 1
∆
= {Y ∈ S n | Y e 1 = 0}
{[ ] [ ] }
0 0
T 0 0
T
= X | X ∈ S n
0 I 0 I
{ [0 √ ] T [ √ ]
= 2VN Z 0 2VN | Z ∈ S
N}
(1878)
[ ] 0 0
T
where P = is an orthogonal projector. Then, similarly, PXP is
0 I
the orthogonal projection of any X ∈ S n on S n 1 in the Euclidean sense (1870),
and
S n⊥
1
{[ ] [ ] }
0 0
T 0 0
T
X − X | X ∈ S n ⊂ S n
0 I 0 I
= { (1879)
ue T 1 + e 1 u T | u∈ R n}
∆
=
Obviously, S n 1 ⊕ S n⊥
1 = S n .
Because {X1 | X ∈ S n } = R n ,
{X − V X V | X ∈ S n } = {1ζ T + ζ1 T − 11 T (1 T ζ 1 n ) | ζ ∈Rn }
= {1ζ T (I − 11 T 1
2n ) + (I − 1
2n 11T )ζ1 T | ζ ∈R n }
where I − 1
2n 11T is invertible.