Is this block matrix invertible? [closed]












1














Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$

Can we prove/disprove that $L$ is invertible?










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closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
    – darij grinberg
    Nov 27 '18 at 21:10
















1














Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$

Can we prove/disprove that $L$ is invertible?










share|cite|improve this question













closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
    – darij grinberg
    Nov 27 '18 at 21:10














1












1








1







Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$

Can we prove/disprove that $L$ is invertible?










share|cite|improve this question













Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$

Can we prove/disprove that $L$ is invertible?







linear-algebra






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asked Nov 27 '18 at 21:01









Math_

113




113




closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
    – darij grinberg
    Nov 27 '18 at 21:10














  • 3




    You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
    – darij grinberg
    Nov 27 '18 at 21:10








3




3




You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10




You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10










3 Answers
3






active

oldest

votes


















1














This is true only when $A$ is square.



Take



$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$



Our matrix $L$ is certainly not invertible.






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    0














    You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
    begin{bmatrix}
    A^{-1} & 0 \
    0 & (A^{T})^{-1}
    end{bmatrix}

    is the inverse of the original matrix.






    share|cite|improve this answer





















    • Here $A$ is not square matrix.
      – Math_
      Nov 27 '18 at 21:15



















    0














    Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.



    A quick aside:




    Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
    (ripped from the wikipedia page on the determinant)




    So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.



    For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
    a & b & 0 & 0 & 0 \
    c & d & 0 & 0 & 0 \
    e & f & 0 & 0 & 0 \
    hline
    0 & 0 & a & c & e \
    0 & 0 & b & d & f \
    end{array} right )$$



    However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.






    share|cite|improve this answer




























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      This is true only when $A$ is square.



      Take



      $$
      A = begin{pmatrix}
      1 \
      0 end{pmatrix}; qquad
      L = begin{pmatrix}
      1&0&0 \
      0&0&0 \
      0&1&0 end{pmatrix}.$$



      Our matrix $L$ is certainly not invertible.






      share|cite|improve this answer


























        1














        This is true only when $A$ is square.



        Take



        $$
        A = begin{pmatrix}
        1 \
        0 end{pmatrix}; qquad
        L = begin{pmatrix}
        1&0&0 \
        0&0&0 \
        0&1&0 end{pmatrix}.$$



        Our matrix $L$ is certainly not invertible.






        share|cite|improve this answer
























          1












          1








          1






          This is true only when $A$ is square.



          Take



          $$
          A = begin{pmatrix}
          1 \
          0 end{pmatrix}; qquad
          L = begin{pmatrix}
          1&0&0 \
          0&0&0 \
          0&1&0 end{pmatrix}.$$



          Our matrix $L$ is certainly not invertible.






          share|cite|improve this answer












          This is true only when $A$ is square.



          Take



          $$
          A = begin{pmatrix}
          1 \
          0 end{pmatrix}; qquad
          L = begin{pmatrix}
          1&0&0 \
          0&0&0 \
          0&1&0 end{pmatrix}.$$



          Our matrix $L$ is certainly not invertible.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 22:03









          Santana Afton

          2,5742629




          2,5742629























              0














              You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
              begin{bmatrix}
              A^{-1} & 0 \
              0 & (A^{T})^{-1}
              end{bmatrix}

              is the inverse of the original matrix.






              share|cite|improve this answer





















              • Here $A$ is not square matrix.
                – Math_
                Nov 27 '18 at 21:15
















              0














              You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
              begin{bmatrix}
              A^{-1} & 0 \
              0 & (A^{T})^{-1}
              end{bmatrix}

              is the inverse of the original matrix.






              share|cite|improve this answer





















              • Here $A$ is not square matrix.
                – Math_
                Nov 27 '18 at 21:15














              0












              0








              0






              You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
              begin{bmatrix}
              A^{-1} & 0 \
              0 & (A^{T})^{-1}
              end{bmatrix}

              is the inverse of the original matrix.






              share|cite|improve this answer












              You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
              begin{bmatrix}
              A^{-1} & 0 \
              0 & (A^{T})^{-1}
              end{bmatrix}

              is the inverse of the original matrix.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 27 '18 at 21:13









              mathnoob

              1,799422




              1,799422












              • Here $A$ is not square matrix.
                – Math_
                Nov 27 '18 at 21:15


















              • Here $A$ is not square matrix.
                – Math_
                Nov 27 '18 at 21:15
















              Here $A$ is not square matrix.
              – Math_
              Nov 27 '18 at 21:15




              Here $A$ is not square matrix.
              – Math_
              Nov 27 '18 at 21:15











              0














              Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.



              A quick aside:




              Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
              (ripped from the wikipedia page on the determinant)




              So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.



              For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
              a & b & 0 & 0 & 0 \
              c & d & 0 & 0 & 0 \
              e & f & 0 & 0 & 0 \
              hline
              0 & 0 & a & c & e \
              0 & 0 & b & d & f \
              end{array} right )$$



              However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.






              share|cite|improve this answer


























                0














                Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.



                A quick aside:




                Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
                (ripped from the wikipedia page on the determinant)




                So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.



                For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
                a & b & 0 & 0 & 0 \
                c & d & 0 & 0 & 0 \
                e & f & 0 & 0 & 0 \
                hline
                0 & 0 & a & c & e \
                0 & 0 & b & d & f \
                end{array} right )$$



                However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.



                  A quick aside:




                  Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
                  (ripped from the wikipedia page on the determinant)




                  So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.



                  For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
                  a & b & 0 & 0 & 0 \
                  c & d & 0 & 0 & 0 \
                  e & f & 0 & 0 & 0 \
                  hline
                  0 & 0 & a & c & e \
                  0 & 0 & b & d & f \
                  end{array} right )$$



                  However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.






                  share|cite|improve this answer












                  Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.



                  A quick aside:




                  Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
                  (ripped from the wikipedia page on the determinant)




                  So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.



                  For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
                  a & b & 0 & 0 & 0 \
                  c & d & 0 & 0 & 0 \
                  e & f & 0 & 0 & 0 \
                  hline
                  0 & 0 & a & c & e \
                  0 & 0 & b & d & f \
                  end{array} right )$$



                  However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 21:32









                  Joe

                  64119




                  64119















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