Exercise:
bf Definition. Let V be a vector space over K and Tin textEndV. Let vin V. Define langle vrangle_T:textSpT^kv| kin mathbbZ_geq subseteq V i.e. langle vrangle_TtextSpvTvT^vT^v.... We call langle vrangle_T the T-cyclic subspace generated by v. abcliste abc langle vrangle_T is invariant under T i.e. Tlangle vrangle_Tsubseteq langle vrangle_T. abc langle vrangle_T is the smallest linear subspace of V which contains v and is T-invariant i.e. if Usubseteq V is a linear subspace s.t. vin U and TUsubseteq U then langle vrangle_Tsubseteq U. abcliste bf Lemma . Suppose that vTv...T^d-v dgeq are linearly indepent and T^dvc_v+c_Tv+...+c_c-T^d-v for some c_...c_d-in K. Then abcliste abc mathcalBvTv...T^d-v forms a basis for langle vrangle_T. abc Denote S:T|_langle vrangle_T i.e. S:langle vrangle_Tlongrightarrow langle vrangle_T is the restriction of T to langle vrangle_T. Then S_mathcalB^mathcalB pmatrix & & & hdots & & c_ & & & hdots & & c_ & & & & vdots & vdots vdots & & ddots & & & vdots vdots & vdots & & ddots & & vdots & & & & & c_d- pmatrixin M_dtimes dK textif d mathcalBv S_mathcalB^mathcalBc_ abcliste bf Lemma . Under the asption of the previous Lemma the characteristic polynomial P_Sx of S is P_Sx-^dx^d-c_d-x^d--...-c_x-c_ textif dgeq P_Sx-x-c_ textif d

Solution:
Proof of Lemma . abcliste abc Denote UtextSpmathcalB. Note that forall element win mathcalB we have Twin U. Indeed Tv TTv...TT^d-vin U and TT^d-vT^dVc_v+...+c_d-T^d-vin U. Longrightarrow TUsubseteq U Longrightarrow U is a T-invariant subspace of V. Clearly Usubseteq langle vrangle_T. But we also have langle vrangle_Tsubseteq U by the previous exercise Longrightarrow Ulangle vrangle_T. By asption the elements of mathcalB are linearly indepent hence they form a basis for U. abc Direct calculation. abcliste Proof of Lemma . S_mathcalB^mathcalB-xId pmatrix -x & & & hdots & & c_ & -x & & hdots & & c_ & & & & vdots & vdots vdots & & ddots & & -x & vdots vdots & vdots & & ddots & & vdots & & & & & c_d--x pmatrix * Induction on d. For d direct calculation. Let dgeq . textdetleftS_mathcalB^mathcalB-xIdright -x textdet pmatrix -x & & & hdots & & c_ & -x & & hdots & & c_ & & & & vdots & vdots vdots & & ddots & & -x & vdots vdots & vdots & & ddots & & vdots & & & & & c_d--x pmatrix+-^dc_textdet pmatrix & -x & & hdots & & & & & vdots & & ddots & & -x & & & ddots & pmatrix matrix d-times d- of the same shape as *. Proof of Cayley-Hamilton. Let V be a finite dimensional vector space over K and Tin textEndV. Let P_TxtextdetT-x id_V be the characteristic polynomial of T. We want to show that P_TT. We'll show that P_TTv forall vin V. Let vin V be any vector w.l.o.g ase vneq . Consider langle vrangle_T. Take the maximal dgeq s.t. vTv...T^d-v are linearly indepent. Then T^dvc_v+c_Tv+...+c_d-T^d-v for some c_...c_d-in K. and textdimlangle vrangle_T d. Put n:textdimV. If d n then ext v Tv...T^d-v to a basis epsilonv Tv...T^d-vw_...w_n-d of V. We have T_epsilon^epsilon leftarray@c|c@ matrix S_mathcalB^mathcalB matrix & * hline matrix matrix & C arrayright Longrightarrow P_TxtextdetT-x id_VP_Sx P_Cx. Consider now the omorphism R:P_TTin textEndV. We have RP_CTcirc P_ST Longrigharrow Rv P_CT P_STv. ** But P_STv-^dT^dv-c_d-T^d-v-...-c_v. So from ** we get P_TTv. All the above holds forall neq vin V recall that neq vin V was arbitrary. Longrightarrow P_TT.