Is the action of taking the inverse distributive in any way? For example, is $(A+B)^{-1}= A^{-1} + B^{-1}?$












0












$begingroup$


Sorry for not formatting properly, I can't seem to get the exponents to show up properly (ex. $A^{-1}$).



Can you distribute the act of taking the inverse over a pair of brackets?



For example, is $(A+B)^{-1} = A^{-1} + B^{-1}?$



That doesn't seem right, but there are some proofs that I can't see being possible without having a place to start, and the only place to start seems to be to do something with the inverse operator (if I can call it an operator). For example: Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$



I've no idea how proofs on that post are getting started; I don't understand their first step. For example:



$(C+DD^T)^{-1} =((I+DD^TC^{-1})C)^{-1}$



Why is the above true, and how did the answerer get there? This is why I'm thinking there's some kind of way to expand an inverse.



It always seems to involve the identity matrix, and I realize that $A^{-1}A=I,;$ but I'm still unable to understand those proofs and how they're expanding the expression out in the way that they are.



Any help whatsoever is greatly appreciated.






linear-algebra matrices inverse







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$endgroup$












  • $begingroup$
    They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 21:44


















0












$begingroup$


Sorry for not formatting properly, I can't seem to get the exponents to show up properly (ex. $A^{-1}$).



Can you distribute the act of taking the inverse over a pair of brackets?



For example, is $(A+B)^{-1} = A^{-1} + B^{-1}?$



That doesn't seem right, but there are some proofs that I can't see being possible without having a place to start, and the only place to start seems to be to do something with the inverse operator (if I can call it an operator). For example: Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$



I've no idea how proofs on that post are getting started; I don't understand their first step. For example:



$(C+DD^T)^{-1} =((I+DD^TC^{-1})C)^{-1}$



Why is the above true, and how did the answerer get there? This is why I'm thinking there's some kind of way to expand an inverse.



It always seems to involve the identity matrix, and I realize that $A^{-1}A=I,;$ but I'm still unable to understand those proofs and how they're expanding the expression out in the way that they are.



Any help whatsoever is greatly appreciated.






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$












  • $begingroup$
    They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 21:44
















0












0








0





$begingroup$


Sorry for not formatting properly, I can't seem to get the exponents to show up properly (ex. $A^{-1}$).



Can you distribute the act of taking the inverse over a pair of brackets?



For example, is $(A+B)^{-1} = A^{-1} + B^{-1}?$



That doesn't seem right, but there are some proofs that I can't see being possible without having a place to start, and the only place to start seems to be to do something with the inverse operator (if I can call it an operator). For example: Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$



I've no idea how proofs on that post are getting started; I don't understand their first step. For example:



$(C+DD^T)^{-1} =((I+DD^TC^{-1})C)^{-1}$



Why is the above true, and how did the answerer get there? This is why I'm thinking there's some kind of way to expand an inverse.



It always seems to involve the identity matrix, and I realize that $A^{-1}A=I,;$ but I'm still unable to understand those proofs and how they're expanding the expression out in the way that they are.



Any help whatsoever is greatly appreciated.






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$




Sorry for not formatting properly, I can't seem to get the exponents to show up properly (ex. $A^{-1}$).



Can you distribute the act of taking the inverse over a pair of brackets?



For example, is $(A+B)^{-1} = A^{-1} + B^{-1}?$



That doesn't seem right, but there are some proofs that I can't see being possible without having a place to start, and the only place to start seems to be to do something with the inverse operator (if I can call it an operator). For example: Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$



I've no idea how proofs on that post are getting started; I don't understand their first step. For example:



$(C+DD^T)^{-1} =((I+DD^TC^{-1})C)^{-1}$



Why is the above true, and how did the answerer get there? This is why I'm thinking there's some kind of way to expand an inverse.



It always seems to involve the identity matrix, and I realize that $A^{-1}A=I,;$ but I'm still unable to understand those proofs and how they're expanding the expression out in the way that they are.



Any help whatsoever is greatly appreciated.






linear-algebra matrices inverse




linear-algebra matrices inverse






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share|cite|improve this question













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edited Dec 16 '18 at 22:07









user376343

3,7883827




3,7883827










asked Dec 16 '18 at 21:40









James RonaldJames Ronald

1257




1257












  • $begingroup$
    They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 21:44




















  • $begingroup$
    They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 21:44


















$begingroup$
They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
$endgroup$
– Yves Daoust
Dec 16 '18 at 21:44






$begingroup$
They use the artifact $A=AB^{-1}B$ and distributivity of multiplication over addition.
$endgroup$
– Yves Daoust
Dec 16 '18 at 21:44












4 Answers
4






active

oldest

votes


















3












$begingroup$

Consider the case of $1times 1$ matrices.



Observe
begin{align}
(1+1)^{-1} neq 1^{-1}+1^{-1}.
end{align}






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $A=I$ and $B=-I$ are $ntimes n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        of course matrices have to be invertible ( but i guess this is given)
        $endgroup$
        – Ömer F. Yi
        Dec 16 '18 at 22:15



















      0












      $begingroup$

      Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,cdots, a_n )\B=diag(b_1,b_2,cdots, b_n )$$The we have $$(A+B)^{-1}=diagleft({1over a_1+b_1},{1over a_2+b_2},cdots ,{1over a_n+b_n}right)$$and $$A^{-1}+B^{-1}=diagleft({1over a_1}+{1over b_1},{1over a_2}+{1over b_2},cdots ,{1over a_n}+{1over b_n}right)$$clearly $${1over a_i+b_i}ne {1over a_i}+{1over b_i}$$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Consider the case of $1times 1$ matrices.



        Observe
        begin{align}
        (1+1)^{-1} neq 1^{-1}+1^{-1}.
        end{align}






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Consider the case of $1times 1$ matrices.



          Observe
          begin{align}
          (1+1)^{-1} neq 1^{-1}+1^{-1}.
          end{align}






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Consider the case of $1times 1$ matrices.



            Observe
            begin{align}
            (1+1)^{-1} neq 1^{-1}+1^{-1}.
            end{align}






            share|cite|improve this answer









            $endgroup$



            Consider the case of $1times 1$ matrices.



            Observe
            begin{align}
            (1+1)^{-1} neq 1^{-1}+1^{-1}.
            end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 21:44









            Jacky ChongJacky Chong

            18.5k21128




            18.5k21128























                0












                $begingroup$

                If $A=I$ and $B=-I$ are $ntimes n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  If $A=I$ and $B=-I$ are $ntimes n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    If $A=I$ and $B=-I$ are $ntimes n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix






                    share|cite|improve this answer









                    $endgroup$



                    If $A=I$ and $B=-I$ are $ntimes n$ matrices, both are invertible and equal their own inverses, but their sum is $0$ matrix







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 16 '18 at 22:10









                    user376343user376343

                    3,7883827




                    3,7883827























                        0












                        $begingroup$

                        Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          of course matrices have to be invertible ( but i guess this is given)
                          $endgroup$
                          – Ömer F. Yi
                          Dec 16 '18 at 22:15
















                        0












                        $begingroup$

                        Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          of course matrices have to be invertible ( but i guess this is given)
                          $endgroup$
                          – Ömer F. Yi
                          Dec 16 '18 at 22:15














                        0












                        0








                        0





                        $begingroup$

                        Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.






                        share|cite|improve this answer









                        $endgroup$



                        Actually you can show it more generally. If you show that matrices with standard matirx-addition do have group structure. and it becomes obvious, as for any group, if a,b are elements of your Group (a*b)^-1 =b^-1 * a^-1.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 16 '18 at 22:15









                        Ömer F. YiÖmer F. Yi

                        114




                        114












                        • $begingroup$
                          of course matrices have to be invertible ( but i guess this is given)
                          $endgroup$
                          – Ömer F. Yi
                          Dec 16 '18 at 22:15


















                        • $begingroup$
                          of course matrices have to be invertible ( but i guess this is given)
                          $endgroup$
                          – Ömer F. Yi
                          Dec 16 '18 at 22:15
















                        $begingroup$
                        of course matrices have to be invertible ( but i guess this is given)
                        $endgroup$
                        – Ömer F. Yi
                        Dec 16 '18 at 22:15




                        $begingroup$
                        of course matrices have to be invertible ( but i guess this is given)
                        $endgroup$
                        – Ömer F. Yi
                        Dec 16 '18 at 22:15











                        0












                        $begingroup$

                        Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,cdots, a_n )\B=diag(b_1,b_2,cdots, b_n )$$The we have $$(A+B)^{-1}=diagleft({1over a_1+b_1},{1over a_2+b_2},cdots ,{1over a_n+b_n}right)$$and $$A^{-1}+B^{-1}=diagleft({1over a_1}+{1over b_1},{1over a_2}+{1over b_2},cdots ,{1over a_n}+{1over b_n}right)$$clearly $${1over a_i+b_i}ne {1over a_i}+{1over b_i}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,cdots, a_n )\B=diag(b_1,b_2,cdots, b_n )$$The we have $$(A+B)^{-1}=diagleft({1over a_1+b_1},{1over a_2+b_2},cdots ,{1over a_n+b_n}right)$$and $$A^{-1}+B^{-1}=diagleft({1over a_1}+{1over b_1},{1over a_2}+{1over b_2},cdots ,{1over a_n}+{1over b_n}right)$$clearly $${1over a_i+b_i}ne {1over a_i}+{1over b_i}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,cdots, a_n )\B=diag(b_1,b_2,cdots, b_n )$$The we have $$(A+B)^{-1}=diagleft({1over a_1+b_1},{1over a_2+b_2},cdots ,{1over a_n+b_n}right)$$and $$A^{-1}+B^{-1}=diagleft({1over a_1}+{1over b_1},{1over a_2}+{1over b_2},cdots ,{1over a_n}+{1over b_n}right)$$clearly $${1over a_i+b_i}ne {1over a_i}+{1over b_i}$$






                            share|cite|improve this answer









                            $endgroup$



                            Even doesn't hold for diagonal case. Consider$$A=diag(a_1,a_2,cdots, a_n )\B=diag(b_1,b_2,cdots, b_n )$$The we have $$(A+B)^{-1}=diagleft({1over a_1+b_1},{1over a_2+b_2},cdots ,{1over a_n+b_n}right)$$and $$A^{-1}+B^{-1}=diagleft({1over a_1}+{1over b_1},{1over a_2}+{1over b_2},cdots ,{1over a_n}+{1over b_n}right)$$clearly $${1over a_i+b_i}ne {1over a_i}+{1over b_i}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 16 '18 at 22:18









                            Mostafa AyazMostafa Ayaz

                            15.6k3939




                            15.6k3939






























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