Question

Problem 5 Let U be an n dimensional vector space and T E L(U,U). Let I denote the identity transformation I(u) = u for each u EU and let 0 denote the zero transformation. Show that there is a natural number N, and constants C1, ..., CN+1 such that C1I + c2T + ... + CN+1TN = 0 (Hint: Given dim(U) = n, what is the dimension of L(U,U)? consider ciI + c2T + ... + Cn+11'" = 0, where C1, ..., Cn+1 are constants and 0 is the zero transformation). If you like, you can think of T as an nxn matrix, I as the identity matrix, and 0 as the zero matrix (by problem 1). Note: This problem is asking you to show that there is some polynomial p(x) = (1 + 2x + ... + Cn+12" such that p(T) = 0. But we can be more specific. If p is the characteristic polynomial for T, then p(T) = 0. See problem 5 in homework 2 for an example of this phenomenon. This is called the Cayley Hamilton theorem.

Answer 1

Answey let {1, 7, T ...... THY is linearly dependent of hast less than nt1 members. clearly there is a smalleal natural no. Nxt THEL (IT T? ... TN-1) and {TT, T ... THI} in linearly independent. In other worde, there are is a smallest natural number N niti dim LIIT, T²... TM) = N. Thus, there are scalars Can C, gg. ... CNH dito TH= Col + GT t. . . . .& CRATHI Now, consider the polynomial m= |- - ....-Cal). we claim that m is the minimas polynomial of T. By construction, metto and in is. monic we must show that the degree of each non zero poly. that aasociater the zero linear operator with t, is at least N. let f = 1 ao, ar... and be a non gero polynomial et f(T1=0 and suppore that UEN. By assumption As It aitt... & autu o The scalar as, a,...... au are not all zero. Therefore by the test for 1.D. { 1, T... Illy is L. D. But IT.... TN-17 ix L.I and uks; so {I, T.... TH} is L.I. But this is contractiction proven that Ncu we conclude that min minimal poly of I .. cilt G T + C ₂T + ... + CAHITNO & dim {LCU,U)} = nxn = n²