Page 336 - 35Linear Algebra
P. 336

336                                                                    Sample Second Midterm


                                   (d) Swapping two rows also leaves the permanent unchanged. The argu-
                                       ment is almost the same as in the previous part, except that we need
                                                                          j          i
                                       only reshuffle two matrix elements M     and M     (in the case where
                                                                          σ(i)      σ(j)
                                       rows i and j were swapped). Then we use the fact that summing over
                                       all permutations σ or over all permutations eσ obtained by swapping a
                                       pair in σ are equivalent operations.


                               5. Firstly, lets call (1) = 1 (the 1 × 1 identity matrix). Then we calculate

                                                         T
                                    T
                                                 T T
                                                                  T T
                                                                                                  T
                                                                                      T
                                                                                T T
                                  H = (I −2XX ) = I −2(XX ) = I −2(X ) X = I −2XX = H ,
                                  which demonstrates the first equality. Now we compute
                                           2
                                                                    T
                                                                                  T
                                                         T
                                                                                           T
                                         H = (I − 2XX )(I − 2XX ) = I − 4XX + 4XX XX             T
                                                    T
                                                              T
                                                                                 T
                                                                    T
                                                                                            T
                                         = I − 4XX + 4X(X X)X = I − 4XX + 4X.1.X = I .
                                  So, since HH = I, we have H −1  = H.
                               6. We know Mv = λv. Hence
                                                                                    2
                                                      2
                                                    M v = MMv = Mλv = λMv = λ v ,
                                  and similarly
                                                                                k
                                                          k
                                                       M v = λM    k−1 v = . . . = λ v .
                                                                              k
                                                            k
                                  So v is an eigenvector of M with eigenvalue λ .
                                  Now let us assume v is an eigenvector of the nilpotent matrix N with eigen-
                                  value λ. Then from above
                                                                         k
                                                                  k
                                                                N v = λ v
                                  but by nilpotence, we also have
                                                                   k
                                                                 N v = 0.
                                                                                                    k
                                          k
                                  Hence λ v = 0 and v (being an eigenvector) cannot vanish. Thus λ = 0
                                  and in turn λ = 0.
                               7. Let us think about the eigenvalue problem Mv = λv. This has solutions
                                  when

                                                        3 − λ      −5       2
                                               0 = det                  = λ − 4 ⇒ λ = ±2 .
                                                            1 −3 − λ
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