Does this property of determinants generalize?












2












$begingroup$


Consider the following $2times 2$ matrix
$$
A
=
begin{pmatrix}
a_{11} & a_{12} \
a_{21} & a_{22}
end{pmatrix}.
$$



Let $Delta = a_{11}a_{22}-a_{21}a_{12}$ be its determinant.



Then $A$ has full rank iff $Delta ne 0.$



I noticed that it holds that
$$
A
begin{pmatrix}
a_{22} \
-a_{21}
end{pmatrix}
=
begin{pmatrix}
a_{11} & a_{12} \
a_{21} & a_{22}
end{pmatrix}
begin{pmatrix}
a_{22} \
-a_{21}
end{pmatrix}
=
begin{pmatrix}
Delta \
0
end{pmatrix}.
$$
That is: A linear combination of the columns of $A$ with coefficients that, up to their signs, consist of the elements of the second row of $A$ yields a vector that consists of a zero and the determinant of $A.$



This immediately yields that $A$ is singular if $Delta=0:$ If all numbers in the second row of $A$ are zero, $A$ is singular. Otherwise we have found a nontrival vector in the kernel.



Question: Is there a similar intuition for the determinant of $ntimes n$ matrices?






linear-algebra matrices determinant







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Consider the following $2times 2$ matrix
    $$
    A
    =
    begin{pmatrix}
    a_{11} & a_{12} \
    a_{21} & a_{22}
    end{pmatrix}.
    $$



    Let $Delta = a_{11}a_{22}-a_{21}a_{12}$ be its determinant.



    Then $A$ has full rank iff $Delta ne 0.$



    I noticed that it holds that
    $$
    A
    begin{pmatrix}
    a_{22} \
    -a_{21}
    end{pmatrix}
    =
    begin{pmatrix}
    a_{11} & a_{12} \
    a_{21} & a_{22}
    end{pmatrix}
    begin{pmatrix}
    a_{22} \
    -a_{21}
    end{pmatrix}
    =
    begin{pmatrix}
    Delta \
    0
    end{pmatrix}.
    $$
    That is: A linear combination of the columns of $A$ with coefficients that, up to their signs, consist of the elements of the second row of $A$ yields a vector that consists of a zero and the determinant of $A.$



    This immediately yields that $A$ is singular if $Delta=0:$ If all numbers in the second row of $A$ are zero, $A$ is singular. Otherwise we have found a nontrival vector in the kernel.



    Question: Is there a similar intuition for the determinant of $ntimes n$ matrices?






    linear-algebra matrices determinant







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Consider the following $2times 2$ matrix
      $$
      A
      =
      begin{pmatrix}
      a_{11} & a_{12} \
      a_{21} & a_{22}
      end{pmatrix}.
      $$



      Let $Delta = a_{11}a_{22}-a_{21}a_{12}$ be its determinant.



      Then $A$ has full rank iff $Delta ne 0.$



      I noticed that it holds that
      $$
      A
      begin{pmatrix}
      a_{22} \
      -a_{21}
      end{pmatrix}
      =
      begin{pmatrix}
      a_{11} & a_{12} \
      a_{21} & a_{22}
      end{pmatrix}
      begin{pmatrix}
      a_{22} \
      -a_{21}
      end{pmatrix}
      =
      begin{pmatrix}
      Delta \
      0
      end{pmatrix}.
      $$
      That is: A linear combination of the columns of $A$ with coefficients that, up to their signs, consist of the elements of the second row of $A$ yields a vector that consists of a zero and the determinant of $A.$



      This immediately yields that $A$ is singular if $Delta=0:$ If all numbers in the second row of $A$ are zero, $A$ is singular. Otherwise we have found a nontrival vector in the kernel.



      Question: Is there a similar intuition for the determinant of $ntimes n$ matrices?






      linear-algebra matrices determinant







      share|cite|improve this question









      $endgroup$




      Consider the following $2times 2$ matrix
      $$
      A
      =
      begin{pmatrix}
      a_{11} & a_{12} \
      a_{21} & a_{22}
      end{pmatrix}.
      $$



      Let $Delta = a_{11}a_{22}-a_{21}a_{12}$ be its determinant.



      Then $A$ has full rank iff $Delta ne 0.$



      I noticed that it holds that
      $$
      A
      begin{pmatrix}
      a_{22} \
      -a_{21}
      end{pmatrix}
      =
      begin{pmatrix}
      a_{11} & a_{12} \
      a_{21} & a_{22}
      end{pmatrix}
      begin{pmatrix}
      a_{22} \
      -a_{21}
      end{pmatrix}
      =
      begin{pmatrix}
      Delta \
      0
      end{pmatrix}.
      $$
      That is: A linear combination of the columns of $A$ with coefficients that, up to their signs, consist of the elements of the second row of $A$ yields a vector that consists of a zero and the determinant of $A.$



      This immediately yields that $A$ is singular if $Delta=0:$ If all numbers in the second row of $A$ are zero, $A$ is singular. Otherwise we have found a nontrival vector in the kernel.



      Question: Is there a similar intuition for the determinant of $ntimes n$ matrices?






      linear-algebra matrices determinant




      linear-algebra matrices determinant






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 18:29









      EpiousiosEpiousios

      1,545622




      1,545622






















          1 Answer
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          active

          oldest

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          4












          $begingroup$

          Yes, that generalizes...but not the way you think, probably.




          1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write
            $$
            A = pmatrix{a & b \c & d}\
            A^+ = pmatrix{d & -c \-b & a}
            $$
            where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = pmatrix{Delta\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.


          2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.







          share|cite|improve this answer









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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Yes, that generalizes...but not the way you think, probably.




            1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write
              $$
              A = pmatrix{a & b \c & d}\
              A^+ = pmatrix{d & -c \-b & a}
              $$
              where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = pmatrix{Delta\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.


            2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.







            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Yes, that generalizes...but not the way you think, probably.




              1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write
                $$
                A = pmatrix{a & b \c & d}\
                A^+ = pmatrix{d & -c \-b & a}
                $$
                where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = pmatrix{Delta\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.


              2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.







              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Yes, that generalizes...but not the way you think, probably.




                1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write
                  $$
                  A = pmatrix{a & b \c & d}\
                  A^+ = pmatrix{d & -c \-b & a}
                  $$
                  where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = pmatrix{Delta\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.


                2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.







                share|cite|improve this answer









                $endgroup$



                Yes, that generalizes...but not the way you think, probably.




                1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write
                  $$
                  A = pmatrix{a & b \c & d}\
                  A^+ = pmatrix{d & -c \-b & a}
                  $$
                  where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = pmatrix{Delta\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.


                2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 18:39









                John HughesJohn Hughes

                63.4k24090




                63.4k24090






























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