v2009.01.01 - Convex Optimization
v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization
586 APPENDIX B. SIMPLE MATRICES Due to symmetry of H , the matrix 2-norm (the spectral norm) is equal to the largest eigenvalue-magnitude. A Householder matrix is thus characterized, H T = H , H −1 = H T , ‖H‖ 2 = 1, H 0 (1529) For example, the permutation matrix ⎡ ⎤ 1 0 0 Ξ = ⎣ 0 0 1 ⎦ (1530) 0 1 0 is a Householder matrix having u=[ 0 1 −1 ] T / √ 2 . Not all permutation matrices are Householder matrices, although all permutation matrices are orthogonal matrices (B.5.1, Ξ T Ξ = I) [287,3.4] because they are made by permuting rows and columns of the identity matrix. Neither are ⎡all symmetric ⎤ permutation matrices Householder matrices; e.g., 0 0 0 1 Ξ = ⎢ 0 0 1 0 ⎥ ⎣ 0 1 0 0 ⎦ (1608) is not a Householder matrix. 1 0 0 0 B.4 Auxiliary V -matrices B.4.1 Auxiliary projector matrix V It is convenient to define a matrix V that arises naturally as a consequence of translating the geometric center α c (5.5.1.0.1) of some list X to the origin. In place of X − α c 1 T we may write XV as in (879) where V = I − 1 N 11T ∈ S N (821) is an elementary matrix called the geometric centering matrix. Any elementary matrix in R N×N has N −1 eigenvalues equal to 1. For the particular elementary matrix V , the N th eigenvalue equals 0. The number of 0 eigenvalues must equal dim N(V ) = 1, by the 0 eigenvalues theorem (A.7.3.0.1), because V =V T is diagonalizable. Because V 1 = 0 (1531)
B.4. AUXILIARY V -MATRICES 587 the nullspace N(V )= R(1) is spanned by the eigenvector 1. The remaining eigenvectors span R(V ) ≡ 1 ⊥ = N(1 T ) that has dimension N −1. Because V 2 = V (1532) and V T = V , elementary matrix V is also a projection matrix (E.3) projecting orthogonally on its range N(1 T ) which is a hyperplane containing the origin in R N V = I − 1(1 T 1) −1 1 T (1533) The {0, 1} eigenvalues also indicate diagonalizable V is a projection matrix. [344,4.1, thm.4.1] Symmetry of V denotes orthogonal projection; from (1791), V T = V , V † = V , ‖V ‖ 2 = 1, V ≽ 0 (1534) Matrix V is also circulant [142]. B.4.1.0.1 Example. Relationship of auxiliary to Householder matrix. Let H ∈ S N be a Householder matrix (1527) defined by ⎡ ⎤ u = ⎢ 1. ⎥ ⎣ 1 1 + √ ⎦ ∈ RN (1535) N Then we have [130,2] Let D ∈ S N h and define [ I 0 V = H 0 T 0 [ −HDH = ∆ A b − b T c ] H (1536) ] (1537) where b is a vector. Then because H is nonsingular (A.3.1.0.5) [158,3] [ ] A 0 −V DV = −H 0 T H ≽ 0 ⇔ −A ≽ 0 (1538) 0 and affine dimension is r = rankA when D is a Euclidean distance matrix.
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586 APPENDIX B. SIMPLE MATRICES<br />
Due to symmetry of H , the matrix 2-norm (the spectral norm) is equal to the<br />
largest eigenvalue-magnitude. A Householder matrix is thus characterized,<br />
H T = H , H −1 = H T , ‖H‖ 2 = 1, H 0 (1529)<br />
For example, the permutation matrix<br />
⎡ ⎤<br />
1 0 0<br />
Ξ = ⎣ 0 0 1 ⎦ (1530)<br />
0 1 0<br />
is a Householder matrix having u=[ 0 1 −1 ] T / √ 2 . Not all permutation<br />
matrices are Householder matrices, although all permutation matrices<br />
are orthogonal matrices (B.5.1, Ξ T Ξ = I) [287,3.4] because they are<br />
made by permuting rows and columns of the identity matrix. Neither<br />
are ⎡all symmetric ⎤ permutation matrices Householder matrices; e.g.,<br />
0 0 0 1<br />
Ξ = ⎢ 0 0 1 0<br />
⎥<br />
⎣ 0 1 0 0 ⎦ (1608) is not a Householder matrix.<br />
1 0 0 0<br />
B.4 Auxiliary V -matrices<br />
B.4.1<br />
Auxiliary projector matrix V<br />
It is convenient to define a matrix V that arises naturally as a consequence of<br />
translating the geometric center α c (5.5.1.0.1) of some list X to the origin.<br />
In place of X − α c 1 T we may write XV as in (879) where<br />
V = I − 1 N 11T ∈ S N (821)<br />
is an elementary matrix called the geometric centering matrix.<br />
Any elementary matrix in R N×N has N −1 eigenvalues equal to 1. For the<br />
particular elementary matrix V , the N th eigenvalue equals 0. The number<br />
of 0 eigenvalues must equal dim N(V ) = 1, by the 0 eigenvalues theorem<br />
(A.7.3.0.1), because V =V T is diagonalizable. Because<br />
V 1 = 0 (1531)