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19 Jan 2005: I AvI,so positive definiteness implies̄v>Av > 0 andv̄>(A B B>)v > 0. ... ForJacobi’s method,A B B> is the same asA except that the signs of the off-diagonal elements arereversed.

19 Jan 2005: by solving. (A B)x(k1) = Bx(k) b, k = 0, 1, 2,. ... If the real symmetric matrices A and A B B> areboth positive definite, then the spectral radius ofH = (A B)1B is strictly less than one.

29 Jan 2005: an R2n2. Proposition 6 The following is true:. B>B C>C = I, (4.3). ... a>p aq = [ b>p c. >p ]. [. bqcq. ]. = (b>p bq c>p cq) = (B. >B C>C)p,q = δp,q.

17 Feb 2005: mf (b, b). xkymk. ei(bκ1aκ2) mf (b, a). xkymk+ ei(aκ1aκ2). mf (a, a). ... s 1, k = 1, 2, 3, 4,. wherev1 = (b, b), v2 = (b, a), v3 = (a, a), v4 = (a, b).

30 Sep 2005: Ξ = {ξ(a) : ξ(a) 6= ξ(b) for a 6= b} (4.3). ... look like figure 7. B B B ( )( ( )1 0 11)0 1.

26 Sep 2005: Let us compute the commutator of twosuch vector fields F(A,a) and F(B,b). ... Thus. [F(A,a),F(B,b). ]= F(C,c), where (C, c) = (AB BA,Ab Ba).