v2010.10.26 - Convex Optimization

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

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610 APPENDIX A. LINEAR ALGEBRAWe can deduce from these, given nonsingular matrix Z and any particulardimensionally[ ]compatible Y : matrix A∈ S M is positive semidefinite if andZTonly ifY T A[Z Y ] is positive semidefinite. In other words, from theCorollary it follows: for dimensionally compatible ZA ≽ 0 ⇔ Z T AZ ≽ 0 and Z T has a left inverseProducts such as Z † Z and ZZ † are symmetric and positive semidefinitealthough, given A ≽ 0, Z † AZ and ZAZ † are neither necessarily symmetricor positive semidefinite.A.3.1.0.6 Theorem. Symmetric projector semidefinite. [20,III][21,6] [222, p.55] For symmetric idempotent matrices P and RP,R ≽ 0P ≽ R ⇔ R(P ) ⊇ R(R) ⇔ N(P ) ⊆ N(R)(1509)Projector P is never positive definite [333,6.5 prob.20] unless it is theidentity matrix.⋄A.3.1.0.7 Theorem. Symmetric positive semidefinite. [202, p.400]Given real matrix Ψ with rank Ψ = 1Ψ ≽ 0 ⇔ Ψ = uu T (1510)where u is some real vector; id est, symmetry is necessary and sufficient forpositive semidefiniteness of a rank-1 matrix.⋄Proof. Any rank-one matrix must have the form Ψ = uv T . (B.1)Suppose Ψ is symmetric; id est, v = u . For all y ∈ R M , y T uu T y ≥ 0.Conversely, suppose uv T is positive semidefinite. We know that can hold ifand only if uv T + vu T ≽ 0 ⇔ for all normalized y ∈ R M , 2y T uv T y ≥ 0 ;but that is possible only if v = u .The same does not hold true for matrices of higher rank, as Example A.2.1.0.1shows.

A.4. SCHUR COMPLEMENT 611A.4 Schur complementConsider Schur-form partitioned matrix G : Given A T = A and C T = C ,then [59][ ] A BG =B T ≽ 0C(1511)⇔ A ≽ 0, B T (I −AA † ) = 0, C −B T A † B ≽ 0⇔ C ≽ 0, B(I −CC † ) = 0, A−BC † B T ≽ 0where A † denotes the Moore-Penrose (pseudo)inverse (E). In the firstinstance, I − AA † is a symmetric projection matrix orthogonally projectingon N(A T ). (1910) It is apparently requiredR(B) ⊥ N(A T ) (1512)which precludes A = 0 when B is any nonzero matrix. Note that A ≻ 0 ⇒A † =A −1 ; thereby, the projection matrix vanishes. Likewise, in the secondinstance, I − CC † projects orthogonally on N(C T ). It is requiredR(B T ) ⊥ N(C T ) (1513)which precludes C =0 for B nonzero. Again, C ≻ 0 ⇒ C † = C −1 . So weget, for A or C nonsingular,[ ] A BG =B T ≽ 0C⇔A ≻ 0, C −B T A −1 B ≽ 0orC ≻ 0, A−BC −1 B T ≽ 0(1514)When A is full-rank then, for all B of compatible dimension, R(B) is inR(A). Likewise, when C is full-rank, R(B T ) is in R(C). Thus the flavor,for A and C nonsingular,[ ] A BG =B T ≻ 0C⇔ A ≻ 0, C −B T A −1 B ≻ 0⇔ C ≻ 0, A−BC −1 B T ≻ 0(1515)where C − B T A −1 B is called the Schur complement of A in G , while theSchur complement of C in G is A − BC −1 B T . [153,4.8]

610 APPENDIX A. LINEAR ALGEBRAWe can deduce from these, given nonsingular matrix Z and any particulardimensionally[ ]compatible Y : matrix A∈ S M is positive semidefinite if andZTonly ifY T A[Z Y ] is positive semidefinite. In other words, from theCorollary it follows: for dimensionally compatible ZA ≽ 0 ⇔ Z T AZ ≽ 0 and Z T has a left inverseProducts such as Z † Z and ZZ † are symmetric and positive semidefinitealthough, given A ≽ 0, Z † AZ and ZAZ † are neither necessarily symmetricor positive semidefinite.A.3.1.0.6 Theorem. Symmetric projector semidefinite. [20,III][21,6] [222, p.55] For symmetric idempotent matrices P and RP,R ≽ 0P ≽ R ⇔ R(P ) ⊇ R(R) ⇔ N(P ) ⊆ N(R)(1509)Projector P is never positive definite [333,6.5 prob.20] unless it is theidentity matrix.⋄A.3.1.0.7 Theorem. Symmetric positive semidefinite. [202, p.400]Given real matrix Ψ with rank Ψ = 1Ψ ≽ 0 ⇔ Ψ = uu T (1510)where u is some real vector; id est, symmetry is necessary and sufficient forpositive semidefiniteness of a rank-1 matrix.⋄Proof. Any rank-one matrix must have the form Ψ = uv T . (B.1)Suppose Ψ is symmetric; id est, v = u . For all y ∈ R M , y T uu T y ≥ 0.Conversely, suppose uv T is positive semidefinite. We know that can hold ifand only if uv T + vu T ≽ 0 ⇔ for all normalized y ∈ R M , 2y T uv T y ≥ 0 ;but that is possible only if v = u .The same does not hold true for matrices of higher rank, as Example A.2.1.0.1shows.

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