Exercise:
Let V be a finite dimensional vector space over mathbbC. Then every Tin textEndV is trigonalizable.

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 Irighttextdet pmatrix lambda_-x & & & * & lambda_-x & & & & ddots & & & &lambda_n-x pmatrix lambda_-x...lambda_n-x. Consider T_A:K^nlongrightarrow K^n T_AuA u. By the induction hypothesis exists a basis mathcalC of K^n s.t. T_A_mathcalC^mathcalC pmatrix lambda_ & & * & ddots & & &lambda_n+ pmatrix. Denote by epsilon the standard basis of K^n. T_A_mathcalC^mathcalCid_mathcalC^epsilon T_A_epsilon^epsilon id_epsilon^mathcalC Q^-AQ. So Q^-AQ pmatrix lambda_ & & * & ddots & & &lambda_n+ pmatrix. Define P: pmatrix & & hdots & & & & vdots & & Q & & & & & pmatrixin M_n+timesn+K. claim. abcliste abc P is invertible and P^- pmatrix & & hdots & & & & vdots & & Q^- & & & & & pmatrix. abc P^-MP pmatrix lambda_ & *' & hdots & *' & lambda_ & & * vdots & & ddots & & & & lambda_n+ pmatrix. abcliste Proof of the claim. Direct calculation. abcliste abc P P^- pmatrix & & hdots & & & & vdots & & Q Q^- & & & & & pmatrix pmatrix & & hdots & & & & vdots & & I_n & & & & & pmatrix and similarly for P^- P. abc P^-MP pmatrix & & hdots & & & & vdots & & Q^- & & & & & pmatrix pmatrix lambda_ & * & hdots & * & & & vdots & & A & & & & & pmatrix pmatrix & & hdots & & & & vdots & & Q & & & & & pmatrix pmatrix & & hdots & & & & vdots & & Q^- & & & & & pmatrix pmatrix lambda_ & *' & hdots & *' & & & vdots & & AQ & & & & & pmatrix pmatrix lambda_ & *' & hdots & *' & & & vdots & & Q^-AQ & & & & & pmatrix pmatrix lambda_ & *' & hdots & *' & lambda_ & & * vdots & & ddots & & & & lambda_n+ pmatrix. This proves the claim. abcliste We continue with the proof of the theorem. Recall we have a basis mathcalB' for V s.t. T_mathcalB'^mathcalB' pmatrix lambda_ & * & hdots & * & & & vdots & & A & & & & pmatrixM . And we found an invertible matrix P s.t. P^-MP pmatrix lambda_ & & & * & lambda_ & & & & ddots & & & & lambda_n+ pmatrix Claim. exists basis mathcalB for V s.t. id_mathcalB'^mathcalBP. Proof of the claim. Write P pmatrix P_ & P_ & hdots & P_k & hdots & P_m vdots & vdots & hdots & vdots & hdots & vdots P_m & P_m & hdots & P_mk & hdots & P_mm pmatrix mtextdimVn+. Write mathcalB'u_u_...u_m. Define v_&:P_u_+P_u_+...+P_mu_m v_&:P_u_+P_u_+...+P_mu_m &vdots v_k&:P_ku_+P_ku_+...+P_mku_m &vdots v_m&:P_mu_+P_mu_+...+P_mmu_m exr. show that mathcalB:v_...v_m is a basis for V. H. Since textdimVm it's enough to show v_...v_m are linearly indepent. Now suppose c_v_+...+c_mv_m ** substitute * ** use the fact that u_...u_m are linearly indepent and obtain that P pmatrix c_ vdots c_m pmatrix pmatrix vdots pmatrix. But P is invertible Longrightarrow pmatrix c_ vdots c_m pmatrix pmatrix vdots pmatrix. . We are done with the claim. We now have T_mathcalB^mathcalBid_mathcalB^mathcalB'T_mathcalB'^mathcalB'id_mathcalB'^mathcalB pmatrix lambda_ & & * & ddots & & &lambda_n+ pmatrix. bf An alternative and more elegant proof of Longleftarrow of the theorem Proof of Longleftarrow. We'll use induction on ntextdimV and also quotient spaces. 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. Now to the proof. n: In this case V is one dimensional and T_mathcalB^mathcalBlambda so T is even diagonalizable. Let ngeq . Ase that Longleftarrow of the theorem holds for all vector spaces of textdimVleq n and every omorphism of such a space. Let V ve 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_Tx Longrightarrow 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 induction hypothesis exists a basis mathcalCz_...z_n+ for V/U s.t. overlineT_mathcalC^mathcalC pmatrix lambda_ & & * & ddots & & & lambdan+ 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_n+v_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_ & * & vdots & vdots & * & lambda_ & * & vdots & * & & & ddots & vdots & & & &lambda_n+ pmatrix. Indeed Tv_lambda_v_ and for leq ileq n+ we have Tv_ioverlineTv_ioverlineTz_ialpha_iz_+alpha_iz_+...+alpha_i-iz_i-+lambda_iz_i for some alpha_i...alpha_i-iin K. Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i+u for some uin U. Write ubeta_i v_. Longrightarrow Tv_ibeta_iv_+alpha_iv_+...+alpha_i-iv_i-+lambda_iv_i. This shows T is trigonalizable. The induction is now complete.