Invertible Matrices within a Matrix












2












$begingroup$


Suppose A, B are invertible matrices of the same size. Show that
$$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.



I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.






linear-algebra matrices inverse block-matrices







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$endgroup$

















    2












    $begingroup$


    Suppose A, B are invertible matrices of the same size. Show that
    $$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.



    I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.






    linear-algebra matrices inverse block-matrices







    share|cite|improve this question









    New contributor




    Bailey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Suppose A, B are invertible matrices of the same size. Show that
      $$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.



      I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.






      linear-algebra matrices inverse block-matrices







      share|cite|improve this question









      New contributor




      Bailey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Suppose A, B are invertible matrices of the same size. Show that
      $$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.



      I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.






      linear-algebra matrices inverse block-matrices




      linear-algebra matrices inverse block-matrices






      share|cite|improve this question









      New contributor




      Bailey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question








      edited 1 hour ago









      Martin Sleziak

      44.8k9118272




      44.8k9118272






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      asked 2 hours ago









      BaileyBailey

      111




      111




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      Bailey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          3 Answers
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          $begingroup$

          You can check directly that
          $$
          M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
          $$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.



            From this you can compute an explicit inverse.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Since $A$ and $B$ are invertible, we have



              $M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                You can check directly that
                $$
                M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
                $$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  You can check directly that
                  $$
                  M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    You can check directly that
                    $$
                    M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    You can check directly that
                    $$
                    M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Martin ArgeramiMartin Argerami

                    126k1182180




                    126k1182180























                        2












                        $begingroup$

                        Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.



                        From this you can compute an explicit inverse.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.



                          From this you can compute an explicit inverse.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.



                            From this you can compute an explicit inverse.






                            share|cite|improve this answer









                            $endgroup$



                            Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.



                            From this you can compute an explicit inverse.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            copper.hatcopper.hat

                            126k559160




                            126k559160























                                1












                                $begingroup$

                                Since $A$ and $B$ are invertible, we have



                                $M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Since $A$ and $B$ are invertible, we have



                                  $M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Since $A$ and $B$ are invertible, we have



                                    $M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Since $A$ and $B$ are invertible, we have



                                    $M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 32 mins ago









                                    FredFred

                                    45.3k1847




                                    45.3k1847






















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