v2010.10.26 - Convex Optimization

B.3. ELEMENTARY MATRIX 641has N −1 eigenvalues equal to 1 corresponding to real eigenvectors thatspan v ⊥ . The remaining eigenvalueλ = 1 − ζv T u (1637)corresponds to eigenvector u . B.6 From [221, App.7.A.26] the determinant:detE = 1 − tr ( ζuv T) = λ (1638)If λ ≠ 0 then E is invertible; [153] (conferB.1.0.1)E −1 = I + ζ λ uvT (1639)Eigenvectors corresponding to 0 eigenvalues belong to N(E) , andthe number of 0 eigenvalues must be at least dim N(E) which, here,can be at most one. (A.7.3.0.1) The nullspace exists, therefore,when λ=0 ; id est, when v T u=1/ζ ; rather, whenever u belongs tohyperplane {z ∈R N | v T z=1/ζ}. Then (when λ=0) elementary matrix Eis a nonorthogonal projector projecting on its range (E 2 =E ,E.1) andN(E)= R(u) ; eigenvector u spans the nullspace when it exists. Byconservation of dimension, dim R(E)=N −dim N(E). It is apparentfrom (1636) that v ⊥ ⊆ R(E) , but dimv ⊥ =N −1. Hence R(E)≡v ⊥ whenthe nullspace exists, and the remaining eigenvectors span it.In summary, when a nontrivial nullspace of E exists,R(E) = N(v T ), N(E) = R(u), v T u = 1/ζ (1640)illustrated in Figure 158, which is opposite to the assignment of subspacesfor a dyad (Figure 156). Otherwise, R(E)= R N .When E = E T , the spectral norm isB.3.1Householder matrix‖E‖ 2 = max{1, |λ|} (1641)An elementary matrix is called a Householder matrix when it has the definingform, for nonzero vector u [159,5.1.2] [153,4.10.1] [331,7.3] [202,2.2]H = I − 2 uuTu T u ∈ SN (1642)B.6 Elementary matrix E is not always diagonalizable because eigenvector u need not beindependent of the others; id est, u∈v ⊥ is possible.