Matrices: $(X^{−1} + Y^{−1})^{−1} $












1














$(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$



Here X, Y, and X + Y are nonsingular.



The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.



Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?






linear-algebra matrices







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    1














    $(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$



    Here X, Y, and X + Y are nonsingular.



    The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.



    Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?






    linear-algebra matrices







    share|cite|improve this question



























      1












      1








      1


      2





      $(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$



      Here X, Y, and X + Y are nonsingular.



      The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.



      Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?






      linear-algebra matrices







      share|cite|improve this question















      $(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$



      Here X, Y, and X + Y are nonsingular.



      The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.



      Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 24 at 12:32









      amWhy

      191k28224439




      191k28224439










      asked Nov 24 at 11:45









      Danielvanheuven

      457




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          3 Answers
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          1














          First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)



          To prove it just multiply by your r.h.s expression



          begin{eqnarray}
          (X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
          &=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
          &=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
          &=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
          &=& 1 - (X + Y)^{-1}(X + Y) + 1 \
          &=& 1
          end{eqnarray}



          As for the second question, the answer is no. Imagine this case



          $$
          X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
          $$



          neither $X$ nor $Y$ are invertible, but $X + Y$ is






          share|cite|improve this answer





























            0














            In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.






            share|cite|improve this answer





























              0














              We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,



              $(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



              $=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



              $=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$



              $=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$



              $=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$



              $=I$



              Similarly, you can postmultiply and do the same.



              For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).



              Hope it is helpful:)






              share|cite|improve this answer























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                3 Answers
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                1














                First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)



                To prove it just multiply by your r.h.s expression



                begin{eqnarray}
                (X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
                &=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
                &=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
                &=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
                &=& 1 - (X + Y)^{-1}(X + Y) + 1 \
                &=& 1
                end{eqnarray}



                As for the second question, the answer is no. Imagine this case



                $$
                X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
                $$



                neither $X$ nor $Y$ are invertible, but $X + Y$ is






                share|cite|improve this answer


























                  1














                  First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)



                  To prove it just multiply by your r.h.s expression



                  begin{eqnarray}
                  (X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
                  &=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
                  &=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
                  &=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
                  &=& 1 - (X + Y)^{-1}(X + Y) + 1 \
                  &=& 1
                  end{eqnarray}



                  As for the second question, the answer is no. Imagine this case



                  $$
                  X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
                  $$



                  neither $X$ nor $Y$ are invertible, but $X + Y$ is






                  share|cite|improve this answer
























                    1












                    1








                    1






                    First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)



                    To prove it just multiply by your r.h.s expression



                    begin{eqnarray}
                    (X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
                    &=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
                    &=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
                    &=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
                    &=& 1 - (X + Y)^{-1}(X + Y) + 1 \
                    &=& 1
                    end{eqnarray}



                    As for the second question, the answer is no. Imagine this case



                    $$
                    X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
                    $$



                    neither $X$ nor $Y$ are invertible, but $X + Y$ is






                    share|cite|improve this answer












                    First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)



                    To prove it just multiply by your r.h.s expression



                    begin{eqnarray}
                    (X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
                    &=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
                    &=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
                    &=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
                    &=& 1 - (X + Y)^{-1}(X + Y) + 1 \
                    &=& 1
                    end{eqnarray}



                    As for the second question, the answer is no. Imagine this case



                    $$
                    X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
                    $$



                    neither $X$ nor $Y$ are invertible, but $X + Y$ is







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 12:15









                    caverac

                    13.4k21029




                    13.4k21029























                        0














                        In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.






                        share|cite|improve this answer


























                          0














                          In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.






                            share|cite|improve this answer












                            In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 12:19









                            J.G.

                            22.4k22035




                            22.4k22035























                                0














                                We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,



                                $(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                $=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                $=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$



                                $=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$



                                $=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$



                                $=I$



                                Similarly, you can postmultiply and do the same.



                                For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).



                                Hope it is helpful:)






                                share|cite|improve this answer




























                                  0














                                  We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,



                                  $(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                  $=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                  $=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$



                                  $=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$



                                  $=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$



                                  $=I$



                                  Similarly, you can postmultiply and do the same.



                                  For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).



                                  Hope it is helpful:)






                                  share|cite|improve this answer


























                                    0












                                    0








                                    0






                                    We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,



                                    $(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                    $=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$



                                    $=I$



                                    Similarly, you can postmultiply and do the same.



                                    For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).



                                    Hope it is helpful:)






                                    share|cite|improve this answer














                                    We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,



                                    $(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                    $=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$



                                    $=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$



                                    $=I$



                                    Similarly, you can postmultiply and do the same.



                                    For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).



                                    Hope it is helpful:)







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 24 at 12:22

























                                    answered Nov 24 at 12:14









                                    Martund

                                    1,351212




                                    1,351212






























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                                        Musa

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                                        Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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