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v2009.01.01 - Convex Optimization

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E.7. ON VECTORIZED MATRICES OF HIGHER RANK 671<br />

E.7.2.0.2 Example. PXP redux & N(V).<br />

Suppose we define a subspace of m ×n matrices, each elemental matrix<br />

having columns constituting a list whose geometric center (5.5.1.0.1) is the<br />

origin in R m :<br />

R m×n<br />

c<br />

∆<br />

= {Y ∈ R m×n | Y 1 = 0}<br />

= {Y ∈ R m×n | N(Y ) ⊇ 1} = {Y ∈ R m×n | R(Y T ) ⊆ N(1 T )}<br />

= {XV | X ∈ R m×n } ⊂ R m×n (1873)<br />

the nonsymmetric geometric center subspace. Further suppose V ∈ S n is<br />

a projection matrix having N(V )= R(1) and R(V ) = N(1 T ). Then linear<br />

mapping T(X)=XV is the orthogonal projection of any X ∈ R m×n on R m×n<br />

c<br />

in the Euclidean (vectorization) sense because V is symmetric, N(XV )⊇1,<br />

and R(VX T )⊆ N(1 T ).<br />

Now suppose we define a subspace of symmetric n ×n matrices each of<br />

whose columns constitute a list having the origin in R n as geometric center,<br />

S n c<br />

∆<br />

= {Y ∈ S n | Y 1 = 0}<br />

= {Y ∈ S n | N(Y ) ⊇ 1} = {Y ∈ S n | R(Y ) ⊆ N(1 T )}<br />

(1874)<br />

the geometric center subspace. Further suppose V ∈ S n is a projection<br />

matrix, the same as before. Then V XV is the orthogonal projection of<br />

any X ∈ S n on S n c in the Euclidean sense (1870) because V is symmetric,<br />

V XV 1=0, and R(V XV )⊆ N(1 T ). Two-sided projection is necessary only<br />

to remain in the ambient symmetric matrix subspace. Then<br />

S n c = {V XV | X ∈ S n } ⊂ S n (1875)<br />

has dim S n c = n(n−1)/2 in isomorphic R n(n+1)/2 . We find its orthogonal<br />

complement as the aggregate of all negative directions of orthogonal<br />

projection on S n c : the translation-invariant subspace (5.5.1.1)<br />

S n⊥<br />

c<br />

∆<br />

= {X − V XV | X ∈ S n } ⊂ S n<br />

= {u1 T + 1u T | u∈ R n }<br />

(1876)<br />

characterized by the doublet u1 T + 1u T (B.2). E.13 Defining the<br />

E.13 Proof.<br />

{X − V X V | X ∈ S n } = {X − (I − 1 n 11T )X(I − 11 T 1 n ) | X ∈ Sn }<br />

= { 1 n 11T X + X11 T 1 n − 1 n 11T X11 T 1 n | X ∈ Sn }

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