v2010.10.26 - Convex Optimization
v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization
626 APPENDIX A. LINEAR ALGEBRAA.7.20 entryIf a positive semidefinite matrix A = [A ij ] ∈ R n×n has a 0 entry A ii on itsmain diagonal, then A ij + A ji = 0 ∀j . [272,1.3.1]Any symmetric positive semidefinite matrix having a 0 entry on its maindiagonal must be 0 along the entire row and column to which that 0 entrybelongs. [159,4.2.8] [202,7.1 prob.2]A.7.3 0 eigenvalues theoremThis theorem is simple, powerful, and widely applicable:A.7.3.0.1 Theorem. Number of 0 eigenvalues.For any matrix A∈ R m×nrank(A) + dim N(A) = n (1583)by conservation of dimension. [202,0.4.4]For any square matrix A∈ R m×m , number of 0 eigenvalues is at leastequal to dim N(A)dim N(A) ≤ number of 0 eigenvalues ≤ m (1584)while all eigenvectors corresponding to those 0 eigenvalues belong to N(A).[331,5.1] A.16For diagonalizable matrix A (A.5), the number of 0 eigenvalues isprecisely dim N(A) while the corresponding eigenvectors span N(A). Realand imaginary parts of the eigenvectors remaining span R(A).A.16 We take as given the well-known fact that the number of 0 eigenvalues cannot be lessthan dimension of the nullspace. We offer an example of the converse:⎡A = ⎢⎣1 0 1 00 0 1 00 0 0 01 0 0 0dim N(A)=2, λ(A)=[0 0 0 1] T ; three eigenvectors in the nullspace but only two areindependent. The right-hand side of (1584) is tight for nonzero matrices; e.g., (B.1) dyaduv T ∈ R m×m has m 0-eigenvalues when u ∈ v ⊥ .⎤⎥⎦
A.7. ZEROS 627(Transpose.)Likewise, for any matrix A∈ R m×nrank(A T ) + dim N(A T ) = m (1585)For any square A∈ R m×m , number of 0 eigenvalues is at least equalto dim N(A T )=dim N(A) while all left-eigenvectors (eigenvectors of A T )corresponding to those 0 eigenvalues belong to N(A T ).For diagonalizable A , number of 0 eigenvalues is precisely dim N(A T )while the corresponding left-eigenvectors span N(A T ). Real and imaginaryparts of the left-eigenvectors remaining span R(A T ).⋄Proof. First we show, for a diagonalizable matrix, the number of 0eigenvalues is precisely the dimension of its nullspace while the eigenvectorscorresponding to those 0 eigenvalues span the nullspace:Any diagonalizable matrix A∈ R m×m must possess a complete set oflinearly independent eigenvectors. If A is full-rank (invertible), then allm=rank(A) eigenvalues are nonzero. [331,5.1]Suppose rank(A)< m . Then dim N(A) = m−rank(A). Thus there isa set of m−rank(A) linearly independent vectors spanning N(A). Eachof those can be an eigenvector associated with a 0 eigenvalue becauseA is diagonalizable ⇔ ∃ m linearly independent eigenvectors. [331,5.2]Eigenvectors of a real matrix corresponding to 0 eigenvalues must be real. A.17Thus A has at least m−rank(A) eigenvalues equal to 0.Now suppose A has more than m−rank(A) eigenvalues equal to 0.Then there are more than m−rank(A) linearly independent eigenvectorsassociated with 0 eigenvalues, and each of those eigenvectors must be inN(A). Thus there are more than m−rank(A) linearly independent vectorsin N(A) ; a contradiction.Diagonalizable A therefore has rank(A) nonzero eigenvalues and exactlym−rank(A) eigenvalues equal to 0 whose corresponding eigenvectorsspan N(A).By similar argument, the left-eigenvectors corresponding to 0 eigenvaluesspan N(A T ).Next we show when A is diagonalizable, the real and imaginary parts ofits eigenvectors (corresponding to nonzero eigenvalues) span R(A) :A.17 Proof. Let ∗ denote complex conjugation. Suppose A=A ∗ and As i =0. Thens i = s ∗ i ⇒ As i =As ∗ i ⇒ As ∗ i =0. Conversely, As∗ i =0 ⇒ As i=As ∗ i ⇒ s i= s ∗ i .
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626 APPENDIX A. LINEAR ALGEBRAA.7.20 entryIf a positive semidefinite matrix A = [A ij ] ∈ R n×n has a 0 entry A ii on itsmain diagonal, then A ij + A ji = 0 ∀j . [272,1.3.1]Any symmetric positive semidefinite matrix having a 0 entry on its maindiagonal must be 0 along the entire row and column to which that 0 entrybelongs. [159,4.2.8] [202,7.1 prob.2]A.7.3 0 eigenvalues theoremThis theorem is simple, powerful, and widely applicable:A.7.3.0.1 Theorem. Number of 0 eigenvalues.For any matrix A∈ R m×nrank(A) + dim N(A) = n (1583)by conservation of dimension. [202,0.4.4]For any square matrix A∈ R m×m , number of 0 eigenvalues is at leastequal to dim N(A)dim N(A) ≤ number of 0 eigenvalues ≤ m (1584)while all eigenvectors corresponding to those 0 eigenvalues belong to N(A).[331,5.1] A.16For diagonalizable matrix A (A.5), the number of 0 eigenvalues isprecisely dim N(A) while the corresponding eigenvectors span N(A). Realand imaginary parts of the eigenvectors remaining span R(A).A.16 We take as given the well-known fact that the number of 0 eigenvalues cannot be lessthan dimension of the nullspace. We offer an example of the converse:⎡A = ⎢⎣1 0 1 00 0 1 00 0 0 01 0 0 0dim N(A)=2, λ(A)=[0 0 0 1] T ; three eigenvectors in the nullspace but only two areindependent. The right-hand side of (1584) is tight for nonzero matrices; e.g., (B.1) dyaduv T ∈ R m×m has m 0-eigenvalues when u ∈ v ⊥ .⎤⎥⎦
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