Math 110 Midterm Exam Professor K A Ribet November 3 2003 Please put away all books calculators electronic games cell phones pagers mp3 players PDAs and other electronic devices You may refer to a single 2 sided sheet of notes Please write your name on each sheet of paper that you turn in don t trust staples to keep your papers together Explain your answers in full English sentences as is customary and appropriate Your paper is your ambassador when it is graded 1 Let V be a vector space over a field F and let v be a vector in V Let T V V be a linear transformation Suppose that T m v 0 for some positive integer m but that T m 1 v is non zero Show that the span of v T v T m 1 v has dimension m We have to show that the vectors v T v T m 1 v are linearly independent Assume the contrary i e that some non trivial linear combination of these vectors vanishes Take a linear combination with the fewest possible terms and write r X ai T i v with r m 1 We must have ar 6 0 because it in the form 0 i 0 otherwise we could re write the sum with fewer terms Also some ai with i r must be non zero because otherwise the whole sum would consist of one term and we d end up with T r v 0 which is contrary to assumption We apply T m r r r 1 X X i m r to the linear combination and obtain 0 ai T v ai T i m r v i 0 i 0 with the latter equality coming from the assumption T m v 0 We thus have a vanishing linear combination with fewer terms than the minimal one that we started with this is a contradiction 2 Let T V V be a linear map on a non zero finite dimensional vector space V over a field F Suppose that the characteristic polynomial of T splits over F into a product of linear factors Show that there is a basis B of V such that T B is upper triangular We prove the statement by induction on the dimension of V it is trivial if dim V 1 The characteristic polynomial of T has a root and thus T has an eigenvector v1 The vectors that are multiples of v1 form a 1 dimensional subspace W of V that is invariant under T in the sense that T W W As we saw in the homework that was due on October 17 T induces a linear map U V W V W whose characteristic polynomial divides that of T By the induction hypothesis there is a basis v 2 v n of V W in which the matrix of U is upper triangular Here we understand that we are choosing vectors v2 vn of V and that the v i are their images vi W in V W The vectors v1 v2 vn form a basis of V in which T is upper triangular 3 Let V be an n dimensional real or complex inner product space Let e1 en be an orthonormal basis of V Suppose that T V V is a linear transformation n n X X 2 and let T V V be the adjoint of T Show that kT ej k kT ej k2 j 1 j 1 If kT vk kT vk for all v V show that kT vk kT vk for all v V For the first part we just compute Say T ej kT ej k2 X aij 2 so that n X j 1 i kT ej k2 n X X aij ei for each i Then i aij 2 The analogues of the i j 1 aij for T are the aji but interchanging i and j and applying a complex conjugation does not change the double sum that we have just computed Hence we n n X X 2 kT ej k2 For the second part we suppose that kT ej k indeed have j 1 j 1 kT vk kT vk for some v V This vector must be non zero we can assume that it has norm 1 by dividing it by its norm This preserves the inequality that we started with Complete the singleton set v to a basis v v1 v2 vn of V and apply the Gram Schmidt process to this basis to obtain an orthogonal basis of V Divide the resulting vectors by their norms to obtain an orthonormal basis e1 en of V The first element of this basis e1 is v We now look at n n X X 2 kT ej k kT ej k2 comparing terms For each j the jth term on the j 1 j 1 left hand side is less than or equal to the jth term on the right hand side On the other hand the first term on the left is strictly less than the first term on the right This is a contradiction the sums cannot turn out to be equal Our hypothesis kT vk kT vk was therefore incorrect so we are forced to conclude that kT vk kT vk for all v H110 last midterm page 2
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