mihir_naik Posted July 7, 2013 Posted July 7, 2013 Dear all, I have a query regarding the Sylvester's Law of Inertia (Congruence Transformation). It says the following (taken from Gilbert Strang---- 4th edition, Page 324) : C^T*A*C has the same number of positive, negative and zero eigenvalues as A, for some nonsingular matrix C. That is, the signs of eigenvalues are preserved by the congruence transformation. Here, I give 2 matrices A and C^T*A*C where the signs of the eigenvalues do not seem to be preserved: A = -10.0000 -0.2000 -0.1500 5.0000 0.1440 0.3600 5.0000 -2.0000 0 -2.5000 1.4400 0 0 1.0000 -3.0000 0 -0.7200 7.2000 -5.0000 0 0 2.5000 0 0 0 -1.0000 0 0 0.7200 0 0 0 -1.5000 0 0 3.6000 Eigenvalues of A = -7.3508, -1.3815, 0, 0, 0, 0.5524 (Note that A is a singular matrix and hence it has zero eigenvalues) C = 0.3162 0 0 0 0 0 0 0.7071 0 0 0 0 0 0 0.5774 0 0 0 0 0 0 0.4472 0 0 0 0 0 0 0.8333 0 0 0 0 0 0 0.3727 C^T*A*C = -0.66332, -0.0000094371,0, 0.0002162, -0.4184-0.2645*i, -0.4184 + 0.2645*i It is to be noted that C^T*A*C has two complex conjugate eigenvalues (and because A is singular, C^TAC is also singular), whereas A has only real eigenvalues. But it can be seen that A has two negative eigenvalues whereas C^T*A*C seems to have 4 negative eigenvalues (including the complex conjugates). Any help on this issue is appreciated. Thanks.
uncool Posted July 7, 2013 Posted July 7, 2013 You may want to check if the theorem is supposed to apply to symmetric matrices; that is where this kind of congruence transformation usually happens (otherwise, usually C^(-1) A C is used). =Uncool-
studiot Posted July 7, 2013 Posted July 7, 2013 (edited) My windows2000 mathcad gets the same eignvalues for A as you do, but different ones for the transformed matrix. I get the three zeros again, your negative -0.66 and the complex conjugate pair. I think it was Euler had a method to make the complex pair equivalent to a two real values. Edited July 7, 2013 by studiot
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