Invertible Matrices within a Matrix












7












$begingroup$


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











$endgroup$

















    7












    $begingroup$


    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











    $endgroup$















      7












      7








      7


      4



      $begingroup$


      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











      $endgroup$




      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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 at 5:44









      Martin Sleziak

      44.7k10119272




      44.7k10119272










      asked Jan 31 at 4:33









      BaileyBailey

      361




      361






















          4 Answers
          4






          active

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          16












          $begingroup$

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






          share|cite|improve this answer









          $endgroup$





















            8












            $begingroup$

            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









            $endgroup$





















              6












              $begingroup$

              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









              $endgroup$





















                3












                $begingroup$

                Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.



                For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as



                $$
                sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
                $$



                where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.



                Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as



                $$
                (sum_i alpha_i a_i | sum_i beta_i b_i)
                $$



                This is the null vector only when we have both



                $$
                sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
                $$



                Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.






                share|cite|improve this answer









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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  16












                  $begingroup$

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






                  share|cite|improve this answer









                  $endgroup$


















                    16












                    $begingroup$

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






                    share|cite|improve this answer









                    $endgroup$
















                      16












                      16








                      16





                      $begingroup$

                      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 Jan 31 at 4:40









                      Martin ArgeramiMartin Argerami

                      127k1182183




                      127k1182183























                          8












                          $begingroup$

                          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









                          $endgroup$


















                            8












                            $begingroup$

                            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









                            $endgroup$
















                              8












                              8








                              8





                              $begingroup$

                              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 Jan 31 at 4:35









                              copper.hatcopper.hat

                              127k559160




                              127k559160























                                  6












                                  $begingroup$

                                  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









                                  $endgroup$


















                                    6












                                    $begingroup$

                                    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









                                    $endgroup$
















                                      6












                                      6








                                      6





                                      $begingroup$

                                      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 Jan 31 at 6:25









                                      FredFred

                                      46.9k1848




                                      46.9k1848























                                          3












                                          $begingroup$

                                          Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.



                                          For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as



                                          $$
                                          sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
                                          $$



                                          where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.



                                          Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as



                                          $$
                                          (sum_i alpha_i a_i | sum_i beta_i b_i)
                                          $$



                                          This is the null vector only when we have both



                                          $$
                                          sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
                                          $$



                                          Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            3












                                            $begingroup$

                                            Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.



                                            For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as



                                            $$
                                            sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
                                            $$



                                            where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.



                                            Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as



                                            $$
                                            (sum_i alpha_i a_i | sum_i beta_i b_i)
                                            $$



                                            This is the null vector only when we have both



                                            $$
                                            sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
                                            $$



                                            Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              3












                                              3








                                              3





                                              $begingroup$

                                              Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.



                                              For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as



                                              $$
                                              sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
                                              $$



                                              where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.



                                              Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as



                                              $$
                                              (sum_i alpha_i a_i | sum_i beta_i b_i)
                                              $$



                                              This is the null vector only when we have both



                                              $$
                                              sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
                                              $$



                                              Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.



                                              For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as



                                              $$
                                              sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
                                              $$



                                              where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.



                                              Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as



                                              $$
                                              (sum_i alpha_i a_i | sum_i beta_i b_i)
                                              $$



                                              This is the null vector only when we have both



                                              $$
                                              sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
                                              $$



                                              Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Jan 31 at 8:55









                                              chichi

                                              1,424813




                                              1,424813






























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