Exercise:
Let V be a finite dimensional vector space over K and Tin textEndV. Then T is triagonizable iff P_Tx splits as a product of linear factors i.e. P_Txlambda-x^l_...lambda_k-x^l_k.

Solution:
Proof of Longrightarrow. If exists a basis mathcalB s.t. T_mathcalB^mathcalB pmatrix lambda_ & & & * & lambda_ & & & & ddots & & & & & lambda_n pmatrix then P_TxtextdetleftT_mathcalB^mathcalB-x Iright textdet pmatrix lambda_-x & & & * & lambda_-x & & & & ddots & & & & & lambda_n-x pmatrix lambda_-x...lambda_n-x. Proof of Longleftarrow. We'll use inductin on ntextdimV and also quotient spaces. bf Recall: If Usubseteq V is a linear subspace we have the quotient space V/U whose elements are equivalence classes v of vin V where vv' if v-v'in U. Recall also that if V is finite dimensional then so is V/U and textdimV/UtextdimV-textdimU. Not to the proof. n: In this case V is -dimensional and T_mathcalB^mathcalBlambda so T is even diagonalizable. Let ngeq . Ase that Longleftarrow of the theorem holds for all vector spaces of textdimleq n and every omorphism of such a space. Let V be an n+-dimensional vector space and Tin textEndV be s.t. P_Tx splits as a product of linear factors. Let lambdain K be a zero of P_TxLongrightarrow exists an eigenvector v_in textKerT-lambda id. Ext v_ to a basis mathcalB_v_w_...w_n+ of V. Then T_mathcalB_^mathcalB_ pmatrix lambda_ & * & * & * & * & * & * vdots& * & * & * & * & * & * pmatrix pmatrix lambda_ & * & hdots & * & & & vdots& & A & & & & pmatrix We have P_Txlambda_-x P_Ax. Define U:textSpv_ and consider V/U. Since TUsubseteq U we obtain a new omorphism overlineT:V/Ulongrightarrow V/U defined by overlineTv:Tv. The vectors w_...w_n+ form a basis for V/U and in this basis overlineT is represented by A. So P_Txlambda_-x P_overlineTx. Since P_Tx splits o a product of linear factors so does also P_overlineTx. By the induction hypothesis exists a basis mathcalCz_...z_n+ for V/U s.t. overlineT_mathcalC^mathcalC pmatrix lambda_ & & & * & & ddots & & & * & lambda_n+ pmatrix Now each element z_i in mathcalC is represented by some w_iin V i.e. z_iv_i. Such a v_i is not unique but we pick one for every i. Define mathcalBv_...v_n+. We claim that mathcalB is a basis for V. Indeed v_...v_n are linearly indepent because if alpha_v_+alpha_v_+...+alpha_nv_n Longrightarrow alpha_v_+...+alpha_n+v_n+-alpha_v_ in U Longrightarrow alpha_v_+...+alpha_n+v_n+ Longrightarrow alpha_z_+...+alpha_n+z_n+ Longrightarrow alpha_...alpha_n+. This shows that v_...v_n+ are linearly indepent. Since textdimVn+ mathcalBv_...v_n+ is a basis for V. We claim that T_mathcalB^mathcalB pmatrix lambda_ & * & * & hdots & * & lambda_ & * & hdots & * vdots & & ddots & &vdots & & & & lambda_n+ pmatrix Indeed Tv_lambda_v_ and for leq ileq n+ we have Tv_ioverlineTv_ioverlineTz_ialppha_iz_+alpha_iz_+...+alpha_i-iz_i-+lambda_iz_iquad alpha_i...alpha_i-i in K Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i+u quad uin U. Write ubeta_i v_. Longrightarrow Tv_ibetaiv_+alpha_iv_+...+alpha_i-iv_i-+lambda_iv_i So T_mathcalB^mathcalB pmatrix lambda_ & beta_ & hdots & beta_i & hdots & beta_n+ & lambda_ & hdots & alpha_i & hdots & * vdots & & ddots & alpha_i-i & & vdots vdots & vdots & & lambda_i & & * vdots & vdots & & & & * vdots & vdots & & vdots & ddots & * & & & & &lambda_n+ pmatrix This shows T is trigonalizable. The induction is now complete.