Information Theory, Inference, and Learning ... - Inference Group

Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.610 C — Some Mathematicswithf (a)R≡ ∂e(a) R∂ɛ ,(C.15)and similar definitions for e (a)Land f (a)L. We define these left-vectors to be rowvectors, so that the ‘transpose’ operation is not needed and can be banished.We are free to constrain the magnitudes of the eigenvectors in whatever waywe please. Each left-eigenvector and each right-eigenvector has an arbitrarymagnitude. The natural constraints to use are as follows. First, we constrainthe inner products with:e (a)L(ɛ)e(a) R(ɛ) = 1, for all a. (C.16)Expanding the eigenvectors in ɛ, equation (C.19) implies(e (a)L(0) + ɛf(a)L+ · · ·)(e (a)R(a)(0) + ɛf + · · ·) = 1, (C.17)from which we can extract the terms in ɛ, which say:e (a)L(0)f(a)R+ f (a)LWe are now free to choose the two constraints:e (a)L(0)f (a)R= 0, f(a)LRe(a) R(0) = 0 (C.18)e(a) R(0) = 0, (C.19)which in the special case of a symmetric matrix correspond to constrainingthe eigenvectors to be of constant length, as defined by the Euclidean norm.OK, now that we have defined our cast of characters, what do the definingequations (C.11) and (C.9) tell us about our Taylor expansions (C.13) and(C.15)? We expand equation (C.11) in ɛ.(H(0)+ɛV+· · ·)(e (a)R(0)+ɛf (a)Identifying the terms of order ɛ, we have:H(0)f (a)RR+· · ·) = (λ(a) (0)+ɛµ (a) +· · ·)(e (a)R+ Ve(a) R(0) = λ(a) (0)f (a)R(0)+ɛf (a)R+· · ·).(C.20)+ µ(a) e (a)R(0). (C.21)We can extract interesting results from this equation by hitting it with e (b)L (0):e (b)(a)L(0)H(0)fR+ e(b) L (0)Ve(a) R⇒ λ (b) e (b)L(0)f(a)Setting b = a we obtainR+ e(b) L(0)Ve(a) R(0) = e(b)L(0)λ(a) (0)f (a)R(0) = λ(a) (0)e (b)L+ µ(a) e (b)L(0)e(a) R (0).(a)(0)fR + µ(a) δ ab . (C.22)e (a)L(0)Ve(a) R (0) = µ(a) . (C.23)Alternatively, choosing b ≠ a, we obtain:e (b)L(0)Ve(a) R[λ (0) = (a) (0) − λ (b) (0)]e (b)L(0)f (a)R(C.24)⇒ e (b)L(a)(0)fR = 1λ (a) (0) − λ (b) (0) e(b) L(0)Ve(a) R(0). (C.25)Now, assuming that the right-eigenvectors {e (b)R (0)}N b=1form a complete basis,we must be able to writef (a)R= ∑ bw b e (b)R (0),(C.26)