Exercise:
Let V be an n-dimensional vectorspace over K and Tin textEndV. The following statements are equivalent: abcliste abc T is diagonalizable. abc exists a basis for V whose elements are eigenvectors of T. abc The characteristic polynomial P_Tx splits o a product of linear factors and for every eigenvalue of T the geometric and algebraic multiplicities are equal. abc _i^k textdim Eig_Tlambda_itextdimV where lambda_...lambda_kin K are the eigenvalues of T. abc Vcong oplus_i^k textEig_Tlambda_i. abcliste

Solution:
Proof of iff . Already proven. Proof of Longrightarrow . Let mathcalB be a basis in which T_mathcalB^mathcalB has a diagonal form. By changing the order of the elements in mathcalB we can arrange that T_mathcalB^mathcalB has the following shape T_mathcalB^mathcalB pmatrix lambda_ & & & & & & & & & mboxHuge & ddots & & & & & & & & & lambda_ & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & & lambda_k & & & & & & & & & & ddots & mboxHuge & & & & & & & & &lambda_k pmatrixtext denoted by Lambda. mathcalBv_^...v_l_^v_^...v_l_^...v_^k...v_l_k^k where lambda_...lambda_k are eigenvalues of Tlambda_ineq lambda_jquad forall ineq j and v_^i...v_l_i^i are eigenvectors of lambda_i forall i. P_TxtextdetleftT_mathcalB^mathcalB-xs Iright lambda_-x^l_... lambda_k-x^l_k. * This shows that P_Tx splits as a product of linear factors. It also shows that lambda_...lambda_k are all the eigenvalues of T. It remains to show m_gT;lambda_im_aT;lambda_i forall i. Clearly v_^i...v_l_i^iin textEig_Tlambda_im_gT;lambda_i. But from * we also have l_im_aT;lambda_i. So m_aT;lambda_ileq m_gT;lambda_i forall i. At the same time we always have m_gT;lambda_ileq m_aT;lambda_i Longrightarrow m_gT;lambda_im_aT;lambda_i forall i The proof also shows that l_im_gT;lambda_im_aT;lambda_i. Proof of Longrightarrow . We ase that P_Txlambda_-x_^l_...lambda_k-x^l_k and m_gT;lambda_im_aT;lambda_i forall i. Clearly lambda_...lambda_k are all the eigenvalues of T. By definition l_im_aT;lambda_i. We have ntextdimV_i^k m_aT;lambda_i _i^k m_gT;lambda_i _i^k textdim Eig_Tlambda_i Proof of Longrightarrow . Recall that textEig_Tlambda_+...+textEig_Tlambda_k is a direct . textdimtextEig_Tlambda_+...+textEig_Tlambda_ktextdimtextEig_Tlambda_+....+textdimtextEig_Tlambda_ textdimV Since textEig_Tlambda_+...+textEig_Tlambda_ksubseteq V is a linear subspace we have textEig_Tlambda_+...+textEig_Tlambda_kV. But textEig_Tlambda_+...+textEig_Tlambda_kcond textEig_Tlambda_oplus...oplus textEig_Tlambda_k. Proof of Longrightarrow . If VcondtextEig_Tlambda_oplus...oplus textEig_Tlambda_k then since textEig_Tlambda_+...+textEig_Tlambda_k is a direct and also a linear subspace of V we must have textEig_Tlambda_+...+textEig_Tlambda_kV. Longrightarrow If we choose a basis v_^i...v_l_i^i for textEig_Tlambda_i forall i l_im_gT;lambda_i then v_^...v_l_^...v_^k...v_l_k^k is a basis of V and all the vectors in it are eigenvectors of T.