Page 239 - Applied Probability
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10. Molecular Phylogeny
                                10. Let A and B be the 2 × 2 real matrices

                                             A =
                                                                             λ
                                                      b
                                                                          1
                                                         a
                                   Show that
                                                       a −b     ,     B =     λ  0    .     225
                                                   cos b − sin b      B      λ  10
                                         A
                                                a
                                        e  = e                  ,    e   = e          .
                                                   sin b  cos b                 11

                                   (Hints: Note that 2 × 2 matrices of the form  a  −b  are isomorphic
                                                                            b  a

                                   to the complex numbers under the correspondence  a  −b  ↔ a + bi.
                                                                                  b  a
                                   For the second case write B = λI + C.)
                                11. Define matrices

                                                      a  0                b  1
                                              A =            ,   B =            .
                                                      1  a                0  b
                                   Show that AB  = BA and that

                                                          11
                                                     a+b
                                          A B
                                          e e   = e
                                                          12

                                                                 10               01
                                         e A+B  = e  a+b  cosh(1)      + sinh(1)         .
                                                                 01               10
                                           A B
                                                                                      A
                                                                                             B
                                   Hence, e e  = e A+B . (Hint: Use Problem 10 to calculate e and e .
                                                                                      2
                                   For e A+B  write A + B =(a + b)I + R with R satisfying R = I.)
                                                  A
                                12. Prove that det(e )= e tr(A) , where tr is the trace function. (Hint:
                                   Since the diagonalizable matrices are dense in the set of matrices
                                   [11], by continuity you may assume that A is diagonalizable.)
                                13. For the nucleotide substitution model of Section 10.5, prove formally
                                   that P(t) has the same pattern for equality of entries as Λ. For exam-
                                                                                   k
                                   ple, p AC (t)= p GC (t). (Hint: Prove by induction that Λ has the same
                                   pattern as Λ. Then note the matrix exponential definition (10.7).)
                                14. For the nucleotide substitution model of Section 10.5, show that Λ
                                   has eigenvalues 0, −(γ+λ+δ+κ), −(α+ +γ+λ), and −(δ+κ+β+σ)
                                   and corresponding right eigenvectors
                                                          t

                                           1 =    1, 1, 1, 1
                                                                t

                                                        c 5  c 5
                                           u =    1, 1, −  , −
                                                        c 2  c 2
                                                    α(c 5 − c 3 )+ κc 2 − (c 5 − c 3 ) − δc 2    t
                                           v  =                  ,               , 1, 1
                                                   δ(c 3 − c 2 ) −  c 5  δ(c 3 − c 2 ) −  c 5
                                                                                       t

                                                      β(c 2 − c 6 )+ λc 5 −σ(c 2 − c 6 ) − γc 5
                                          w   =   1, 1,              ,                  ,
                                                       γ(c 6 − c 5 ) − σc 2  γ(c 6 − c 5 ) − σc 2
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