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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.

21 May 2005: BA. B. B. B. A A D. A. DCC. AA A A. ... 3. (b) Now show that γ is an epimorphism (i.e., is surjective).

9 Aug 2005: N$9%9 #%>7#,- %>.- "%-. r29B!7#.- >!$&9B&B/ <B&!-B/a -.I!& 7#. -$& , ,-7#Gp! ... 213%#B>! @}(|B&B&29 s BC!,p & R7#. $.p#{S B&B&! F '- ,$7#B&.-,{S = 4$&5.

19 Apr 2005: Equilibrium un-folding data were collected under the same conditions using aPerkin-Elmer LS-50b fluorimeter with excitation and detection at280 nm and 356 nm, respectively (K.T., B.B.K., ... B.B.K., F.M.P., and K.T. acknowledge sup-port from the John and

7 Jul 2005: φ. Section B-B. sd. p. Geometry 1. 10 mm. Geometry 2. ... Geometry 2,G = 0:092: (a) unconstrained compression; (b) bonded compression; (c) constrained compression.

9 Aug 2005: YL < :k| K2 -J4)? > :B,B"E77)W(& 8-B:I; D 7? ... 7! bP : k2q5),) ) &y ,4/0? x :B,B -,)4sE: )E,B4 j H.

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.

21 May 2005: A B = {a b : a A, b B}, AB = {ab : a A, b B}. ... It shouldbe clear that. b. af > 0, and you should explain why this is so.

24 Apr 2005: 7 @?A% B DGF H J /! 78+#/! ; & 5$ ),H ; 3 £? V BN 8% $! ; ' ... N ¥ A "p ), ( " N ¥ A = & Q' 5 B u p 2@A u.

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