Exercise:
Let VW be finite dimensional vector spaces over K mathcalBv_...v_n a basis for V and mathcalCw_...w_n a basis for W. Then Psi_mathcalC^mathcalB:textHomVWlongrightarrow M_mtimes nK defined by Tlongmapsto T_mathcalC^mathcalB:pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & & vdots pmatrix. is a linear map and moreover an isomorphism.

Solution:
Proof. Recall that T_T_mathcalC^mathcalBPsi_mathcalCcirc Tcirc Psi_mathcalB^-. Recall also that Psi_mathcalBvv_mathcalBin K^n forall vin V Psi_mathcalCww_mathcalCin K^m forall win W Psi_mathcalB^-leftarrayc a_ vdots a_n arrayright a_v_+...+a_nv_n Psi_mathcalC^-leftarrayc b_ vdots b_n arrayright b_w_+...+b_mw_m We claim that the map Psi_mathcalC^mathcalB is bijective. Indeed define mathcalH:M_mtimes nKlongrightarrow textHomVW forall Ain M_mtimes nK mathcalHA:Psi_mathcalC^-circ T_Acirc Psi_mathcalB in textHomVW. We have Psi_mathcalC^mathcalBcirc mathcalHA Psi_mathcalC^mathcalBPsi_mathcalC^-circ T_A circ Psi_mathcalB pmatrix vdots & & vdots Psi_mathcalC^-circ T_A circ Psi_mathcalBv__mathcalC & hdots & Psi_mathcalC^-circ T_A circ Psi_mathcalBv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots T_A circ Psi_mathcalBv_ & hdots & T_A circ Psi_mathcalBv_n vdots & &vdots pmatrix pmatrix vdots & & vdots T_Ae_ & hdots & T_Ae_n vdots & &vdots pmatrix A. This proves that Psi_mathcalC^mathcalBcirc mathcalHtextid_M_mtimes nK. Now mathcalHcirc Psi_mathcalC^mathcalBPhi_mathcalC^-circ T_T_mathcalC^mathcalBcircPhi_mathcalB T This shows that mathcalHcirc Psi_mathcalC^mathcalBtextid_textHomVW. This proves that Phi_mathcalC^mathcalB is bijective. It remains to prove that Psi_mathcalC^mathcalB is linear. Indeed let alphain K Tin textHomVW. Then Psi_mathcalC^mathcalBalpha T pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix alpha pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & &vdots pmatrix alpha Psi_mathcalC^mathcalBT Similarly one shows that Psi_mathcalC^mathcalBT_+T_Psi_mathcalC^mathcalBT_+Psi_mathcalC^mathcalBT_.