v2010.10.26 - Convex Optimization

v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization

convexoptimization.com
from convexoptimization.com More from this publisher
12.07.2015 Views

56 CHAPTER 2. CONVEX GEOMETRYConcurrently, consider injective linear operator Py=A † y : R m → R nwhere R(A † )= R(A T ). P(Ax)= PTx achieves projection of vector x onthe row space R(A T ). (E.3.1) This means vector Ax can be succinctlyinterpreted as coefficients of orthogonal projection.Pseudoinverse matrix A † is skinny and full-rank, so operator Py is a linearbijection with respect to its range R(A † ). By Definition 2.2.1.0.1, imageP(T B) of projection PT(B) on R(A T ) in R n must therefore be isomorphicwith the set of projection coefficients T B = {Ax |x∈ B} in R m and have thesame affine dimension by (49). To illustrate, we present a three-dimensionalEuclidean body B in Figure 19 where any point x in the nullspace N(A)maps to the origin.2.2.2 Symmetric matrices2.2.2.0.1 Definition. Symmetric matrix subspace.Define a subspace of R M×M : the convex set of all symmetric M×M matrices;S M { A∈ R M×M | A=A T} ⊆ R M×M (50)This subspace comprising symmetric matrices S M is isomorphic with thevector space R M(M+1)/2 whose dimension is the number of free variables in asymmetric M ×M matrix. The orthogonal complement [331] [250] of S M isS M⊥ { A∈ R M×M | A=−A T} ⊂ R M×M (51)the subspace of antisymmetric matrices in R M×M ; id est,S M ⊕ S M⊥ = R M×M (52)where unique vector sum ⊕ is defined on page 772.△All antisymmetric matrices are hollow by definition (have 0 maindiagonal). Any square matrix A∈ R M×M can be written as a sum of itssymmetric and antisymmetric parts: respectively,A = 1 2 (A +AT ) + 1 2 (A −AT ) (53)The symmetric part is orthogonal in R M2 to the antisymmetric part; videlicet,tr ( (A +A T )(A −A T ) ) = 0 (54)

2.2. VECTORIZED-MATRIX INNER PRODUCT 57In the ambient space of real matrices, the antisymmetric matrix subspacecan be described{ }1S M⊥ =2 (A −AT ) | A∈ R M×M ⊂ R M×M (55)because any matrix in S M is orthogonal to any matrix in S M⊥ . Furtherconfined to the ambient subspace of symmetric matrices, because ofantisymmetry, S M⊥ would become trivial.2.2.2.1 Isomorphism of symmetric matrix subspaceWhen a matrix is symmetric in S M , we may still employ the vectorizationtransformation (37) to R M2 ; vec , an isometric isomorphism. We mightinstead choose to realize in the lower-dimensional subspace R M(M+1)/2 byignoring redundant entries (below the main diagonal) during transformation.Such a realization would remain isomorphic but not isometric. Lack ofisometry is a spatial distortion due now to disparity in metric between R M2and R M(M+1)/2 . To realize isometrically in R M(M+1)/2 , we must make acorrection: For Y = [Y ij ]∈ S M we take symmetric vectorization [215,2.2.1]⎡ ⎤√2Y12Y 11Y 22√2Y13svec Y √2Y23∈ R M(M+1)/2 (56)⎢ Y 33 ⎥⎣ ⎦.Y MMwhere all entries off the main diagonal have been scaled. Now for Z ∈ S M〈Y , Z〉 tr(Y T Z) = vec(Y ) T vec Z = svec(Y ) T svec Z (57)Then because the metrics become equivalent, for X ∈ S M‖ svec X − svec Y ‖ 2 = ‖X − Y ‖ F (58)and because symmetric vectorization (56) is a linear bijective mapping, thensvec is an isometric isomorphism of the symmetric matrix subspace. In other

2.2. VECTORIZED-MATRIX INNER PRODUCT 57In the ambient space of real matrices, the antisymmetric matrix subspacecan be described{ }1S M⊥ =2 (A −AT ) | A∈ R M×M ⊂ R M×M (55)because any matrix in S M is orthogonal to any matrix in S M⊥ . Furtherconfined to the ambient subspace of symmetric matrices, because ofantisymmetry, S M⊥ would become trivial.2.2.2.1 Isomorphism of symmetric matrix subspaceWhen a matrix is symmetric in S M , we may still employ the vectorizationtransformation (37) to R M2 ; vec , an isometric isomorphism. We mightinstead choose to realize in the lower-dimensional subspace R M(M+1)/2 byignoring redundant entries (below the main diagonal) during transformation.Such a realization would remain isomorphic but not isometric. Lack ofisometry is a spatial distortion due now to disparity in metric between R M2and R M(M+1)/2 . To realize isometrically in R M(M+1)/2 , we must make acorrection: For Y = [Y ij ]∈ S M we take symmetric vectorization [215,2.2.1]⎡ ⎤√2Y12Y 11Y 22√2Y13svec Y √2Y23∈ R M(M+1)/2 (56)⎢ Y 33 ⎥⎣ ⎦.Y MMwhere all entries off the main diagonal have been scaled. Now for Z ∈ S M〈Y , Z〉 tr(Y T Z) = vec(Y ) T vec Z = svec(Y ) T svec Z (57)Then because the metrics become equivalent, for X ∈ S M‖ svec X − svec Y ‖ 2 = ‖X − Y ‖ F (58)and because symmetric vectorization (56) is a linear bijective mapping, thensvec is an isometric isomorphism of the symmetric matrix subspace. In other

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!