f(AB), f(BA)  Symmetr’n f(Jordan block) Sign function
                        Five Theorems in Matrix Analysis,
                                           with Applications
                                                   Nick Higham
                                           School of Mathematics
                                     TheUniversity of Manchester
                                        higham@ma.man.ac.uk
                        http://www.ma.man.ac.uk/~higham/
                                 Dundee(EMS)—March17,2006
                                                 Nick Higham       Matrix Analysis                                        1
       f(AB), f(BA)  Symmetr’n f(Jordan block) Sign function       WMFME Λ(AB)andΛ(BA) f(αI +AB)
   Outline
          f(AB) and f(BA)
                 WMFME
                 Λ(AB)andΛ(BA)
                 f(αI +AB)
          Symmetrization
          Jordan Structure of f(A)
          Matrix Sign Identities
                                                 Nick Higham       Matrix Analysis                                        2
       f(AB), f(BA)  Symmetr’n f(Jordan block) Sign function       WMFME Λ(AB)andΛ(BA) f(αI +AB)
   f(AB) and f(BA)
          For A;B ∈ Cn×n, AB 6= BA.
                   HowareABandBArelated?
                   Howaref(AB)andf(BA)related?
                                                              m×n                 n×m
                   Samequestionif A ∈ C                             , B ∈ C             .
                   Generalize to f(αI +AB) and f(αI +BA).
                                                    m                              n
                                                 Nick Higham       Matrix Analysis                                        3
       f(AB), f(BA)  Symmetr’n f(Jordan block) Sign function       WMFME Λ(AB)andΛ(BA) f(αI +AB)
   Sherman–Morrison–Woodbury Formula
          If U;V ∈ Cn×p and I +V∗A−1U is nonsingular then
                                               p
                                 ∗ −1           −1          −1                   ∗   −1       −1 ∗ −1
                 (A+UV ) =A −A U(Ip+V A U) V A :
          Obtained, using A+UV∗ = A(I +A−1U ·V∗), from its
          simpler version
                                      −1                                  −1           A∈Cm×n
                    (Im +AB)               =I−A(In+BA) B                                  B∈Cn×m
                                                 Nick Higham       Matrix Analysis                                        4