Differentiability of $L_A(X)=AX$

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Let $G$ be a matrix group and $A in G$. Define $L_A(X)= AX : G to G$.



Show that



(1) $L_A$ is differentiable. Compute its derivative.



(2) $d(L_A)_I : T_I(G) to T_A(G)$ is an isomorphism. [ Clearly, $L_A$ is invertible and it has a differentiable inverse by (1)]




linear-algebra linear-transformations differential-operators




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up vote
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down vote

favorite












Let $G$ be a matrix group and $A in G$. Define $L_A(X)= AX : G to G$.



Show that



(1) $L_A$ is differentiable. Compute its derivative.



(2) $d(L_A)_I : T_I(G) to T_A(G)$ is an isomorphism. [ Clearly, $L_A$ is invertible and it has a differentiable inverse by (1)]




linear-algebra linear-transformations differential-operators




share|cite|improve this question

















  • 1




    Can you detail your thoughts and what you’ve done?
    – mathcounterexamples.net
    Jul 22 at 6:04












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $G$ be a matrix group and $A in G$. Define $L_A(X)= AX : G to G$.



Show that



(1) $L_A$ is differentiable. Compute its derivative.



(2) $d(L_A)_I : T_I(G) to T_A(G)$ is an isomorphism. [ Clearly, $L_A$ is invertible and it has a differentiable inverse by (1)]




linear-algebra linear-transformations differential-operators




share|cite|improve this question













Let $G$ be a matrix group and $A in G$. Define $L_A(X)= AX : G to G$.



Show that



(1) $L_A$ is differentiable. Compute its derivative.



(2) $d(L_A)_I : T_I(G) to T_A(G)$ is an isomorphism. [ Clearly, $L_A$ is invertible and it has a differentiable inverse by (1)]





linear-algebra linear-transformations differential-operators





share|cite|improve this question












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edited Jul 22 at 8:19









paf

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asked Jul 22 at 5:35









ChakSayantan

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596







  • 1




    Can you detail your thoughts and what you’ve done?
    – mathcounterexamples.net
    Jul 22 at 6:04












  • 1




    Can you detail your thoughts and what you’ve done?
    – mathcounterexamples.net
    Jul 22 at 6:04







1




1




Can you detail your thoughts and what you’ve done?
– mathcounterexamples.net
Jul 22 at 6:04




Can you detail your thoughts and what you’ve done?
– mathcounterexamples.net
Jul 22 at 6:04










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The differential is given by $d(L_A)_X H = AH$.



$$lim_Hto 0fracL_A(X+H) - L_A(X) - AH = lim_Hto 0frac = 0$$



So $d(L_A)_I$ is invertible, with the inverse being $H mapsto A^-1H$.






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    The differential is given by $d(L_A)_X H = AH$.



    $$lim_Hto 0fracL_A(X+H) - L_A(X) - AH = lim_Hto 0frac = 0$$



    So $d(L_A)_I$ is invertible, with the inverse being $H mapsto A^-1H$.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      The differential is given by $d(L_A)_X H = AH$.



      $$lim_Hto 0fracL_A(X+H) - L_A(X) - AH = lim_Hto 0frac = 0$$



      So $d(L_A)_I$ is invertible, with the inverse being $H mapsto A^-1H$.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The differential is given by $d(L_A)_X H = AH$.



        $$lim_Hto 0fracL_A(X+H) - L_A(X) - AH = lim_Hto 0frac = 0$$



        So $d(L_A)_I$ is invertible, with the inverse being $H mapsto A^-1H$.






        share|cite|improve this answer













        The differential is given by $d(L_A)_X H = AH$.



        $$lim_Hto 0fracL_A(X+H) - L_A(X) - AH = lim_Hto 0frac = 0$$



        So $d(L_A)_I$ is invertible, with the inverse being $H mapsto A^-1H$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 22 at 11:40









        mechanodroid

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