v2007.09.13 - Convex Optimization
v2007.09.13 - Convex Optimization v2007.09.13 - Convex Optimization
500 APPENDIX A. LINEAR ALGEBRAA.4.0.0.3 Theorem. Rank of partitioned matrices.When symmetric matrix A is invertible and C is symmetric,[ ] [ ]A B A 0rankB T = rankC 0 T C −B T A −1 B= rankA + rank(C −B T A −1 B)(1320)equals rank of a block on the main diagonal plus rank of its Schur complement[298,2.2, prob.7]. Similarly, when symmetric matrix C is invertible and Ais symmetric,[ ] [ ]A B A − BCrankB T = rank−1 B T 0C0 T C(1321)= rank(A − BC −1 B T ) + rankCProof. The first assertion (1320) holds if and only if [149,0.4.6(c)][ ] [ ]A B A 0∃ nonsingular X,Y XB T Y =C 0 T C −B T A −1 (1322)BLet [149,7.7.6]Y = X T =[ I −A −1 B0 T I]⋄(1323)A.4.0.0.4 Lemma. Rank of Schur-form block. [91] [89]Matrix B ∈ R m×n has rankB≤ ρ if and only if there exist matrices A∈ S mand C ∈ S n such that[ ] A BrankA + rankC ≤ 2ρ andB T ≽ 0 (1324)C⋄A.4.1Determinant[ ] A BG =B T C(1325)We consider again a matrix G partitioned similarly to (1307), but notnecessarily positive (semi)definite, where A and C are symmetric.
A.4. SCHUR COMPLEMENT 501When A is invertible,detG = detA det(C − B T A −1 B) (1326)When C is invertible,detG = detC det(A − BC −1 B T ) (1327)When B is full-rank and skinny, C = 0, and A ≽ 0, then [46,10.1.1]detG ≠ 0 ⇔ A + BB T ≻ 0 (1328)When B is a (column) vector, then for all C ∈ R and all A of dimensioncompatible with Gwhile for C ≠ 0detG = det(A)C − B T A T cofB (1329)detG = C det(A − 1 C BBT ) (1330)where A cof is the matrix of cofactors [247,4] corresponding to A .When B is full-rank and fat, A = 0, and C ≽ 0, thendetG ≠ 0 ⇔ C + B T B ≻ 0 (1331)When B is a row vector, then for A ≠ 0 and all C of dimensioncompatible with GdetG = A det(C − 1 A BT B) (1332)while for all A∈ RdetG = det(C)A − BCcofB T T (1333)where C cof is the matrix of cofactors corresponding to C .
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500 APPENDIX A. LINEAR ALGEBRAA.4.0.0.3 Theorem. Rank of partitioned matrices.When symmetric matrix A is invertible and C is symmetric,[ ] [ ]A B A 0rankB T = rankC 0 T C −B T A −1 B= rankA + rank(C −B T A −1 B)(1320)equals rank of a block on the main diagonal plus rank of its Schur complement[298,2.2, prob.7]. Similarly, when symmetric matrix C is invertible and Ais symmetric,[ ] [ ]A B A − BCrankB T = rank−1 B T 0C0 T C(1321)= rank(A − BC −1 B T ) + rankCProof. The first assertion (1320) holds if and only if [149,0.4.6(c)][ ] [ ]A B A 0∃ nonsingular X,Y XB T Y =C 0 T C −B T A −1 (1322)BLet [149,7.7.6]Y = X T =[ I −A −1 B0 T I]⋄(1323)A.4.0.0.4 Lemma. Rank of Schur-form block. [91] [89]Matrix B ∈ R m×n has rankB≤ ρ if and only if there exist matrices A∈ S mand C ∈ S n such that[ ] A BrankA + rankC ≤ 2ρ andB T ≽ 0 (1324)C⋄A.4.1Determinant[ ] A BG =B T C(1325)We consider again a matrix G partitioned similarly to (1307), but notnecessarily positive (semi)definite, where A and C are symmetric.
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