v2010.10.26 - Convex Optimization

598 APPENDIX A. LINEAR ALGEBRAbecause nonHermitian matrices could be regarded positive semidefinite,rather because nonHermitian (includes nonsymmetric real) matrices are notcomparable on the real line under x H Ax . Yet that complex edifice isdismantled in the test of real matrices (1431) because the domain of testis no longer necessarily complex; meaning, x T Ax will certainly always bereal, regardless of symmetry, and so real A will always be comparable.In summary, if we limit the domain of test to all x in R n as in (1431),then nonsymmetric real matrices are admitted to the realm of semidefinitematrices because they become comparable on the real line. One importantexception occurs for rank-one matrices Ψ=uv T where u and v are realvectors: Ψ is positive semidefinite if and only if Ψ=uu T . (A.3.1.0.7)We might choose to expand the domain of test to all x in C n so that onlysymmetric matrices would be comparable. The alternative to expandingdomain of test is to assume all matrices of interest to be symmetric; thatis commonly done, hence the synonymous relationship with semidefinitematrices.A.2.1.0.1 Example. Nonsymmetric positive definite product.Horn & Johnson assert and Zhang agrees:If A,B ∈ C n×n are positive definite, then we know that theproduct AB is positive definite if and only if AB is Hermitian.[202,7.6 prob.10] [393,6.2,3.2]Implicitly in their statement, A and B are assumed individually Hermitianand the domain of test is assumed complex.We prove that assertion to be false for real matrices under (1431) thatadopts a real domain of test.A T = A =⎡⎢⎣3 0 −1 00 5 1 0−1 1 4 10 0 1 4⎤⎥⎦ ,⎡λ(A) = ⎢⎣5.94.53.42.0⎤⎥⎦ (1436)B T = B =⎡⎢⎣4 4 −1 −14 5 0 0−1 0 5 1−1 0 1 4⎤⎥⎦ ,⎡λ(B) = ⎢⎣8.85.53.30.24⎤⎥⎦ (1437)