trace and determinant of backward diagonal matrix












0












$begingroup$


I am just wondering whether there is any general formula or relation between trace of backward / anti diagonal matrix just like we have in a diagonal matrix of order $n*n$ where trace is the sum of diagonal elements and determinant is the product of diagonal elements.






matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is interesting
    $endgroup$
    – Cloud JR
    Dec 21 '18 at 17:16












  • $begingroup$
    The trace is not the product of the diagonal elements...
    $endgroup$
    – SvanN
    Dec 21 '18 at 17:20






  • 1




    $begingroup$
    Here is another post about antidiagonal matrices that you might find interesting
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:26
















0












$begingroup$


I am just wondering whether there is any general formula or relation between trace of backward / anti diagonal matrix just like we have in a diagonal matrix of order $n*n$ where trace is the sum of diagonal elements and determinant is the product of diagonal elements.






matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is interesting
    $endgroup$
    – Cloud JR
    Dec 21 '18 at 17:16












  • $begingroup$
    The trace is not the product of the diagonal elements...
    $endgroup$
    – SvanN
    Dec 21 '18 at 17:20






  • 1




    $begingroup$
    Here is another post about antidiagonal matrices that you might find interesting
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:26














0












0








0





$begingroup$


I am just wondering whether there is any general formula or relation between trace of backward / anti diagonal matrix just like we have in a diagonal matrix of order $n*n$ where trace is the sum of diagonal elements and determinant is the product of diagonal elements.






matrices







share|cite|improve this question











$endgroup$




I am just wondering whether there is any general formula or relation between trace of backward / anti diagonal matrix just like we have in a diagonal matrix of order $n*n$ where trace is the sum of diagonal elements and determinant is the product of diagonal elements.






matrices




matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:51









Bernard

123k741117




123k741117










asked Dec 21 '18 at 17:11









sejysejy

1589




1589












  • $begingroup$
    This is interesting
    $endgroup$
    – Cloud JR
    Dec 21 '18 at 17:16












  • $begingroup$
    The trace is not the product of the diagonal elements...
    $endgroup$
    – SvanN
    Dec 21 '18 at 17:20






  • 1




    $begingroup$
    Here is another post about antidiagonal matrices that you might find interesting
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:26


















  • $begingroup$
    This is interesting
    $endgroup$
    – Cloud JR
    Dec 21 '18 at 17:16












  • $begingroup$
    The trace is not the product of the diagonal elements...
    $endgroup$
    – SvanN
    Dec 21 '18 at 17:20






  • 1




    $begingroup$
    Here is another post about antidiagonal matrices that you might find interesting
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:26
















$begingroup$
This is interesting
$endgroup$
– Cloud JR
Dec 21 '18 at 17:16






$begingroup$
This is interesting
$endgroup$
– Cloud JR
Dec 21 '18 at 17:16














$begingroup$
The trace is not the product of the diagonal elements...
$endgroup$
– SvanN
Dec 21 '18 at 17:20




$begingroup$
The trace is not the product of the diagonal elements...
$endgroup$
– SvanN
Dec 21 '18 at 17:20




1




1




$begingroup$
Here is another post about antidiagonal matrices that you might find interesting
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:26




$begingroup$
Here is another post about antidiagonal matrices that you might find interesting
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:26










2 Answers
2






active

oldest

votes


















2












$begingroup$

The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $n$ is even, then the trace is zero since all diagonal elements are zero. If $n$ is odd, then the $frac{n+1}{2}, frac{n+1}{2}$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.



The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $-1$ depending on the value of $n$ (specifically, if $n equiv 2 mod 4$ or $n equiv 3 mod 4$ then you need to multiply by $-1$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Take
    $$
    M = pmatrix{&&a_1\&cdots\a_n}
    $$
    to be your anti-diagonal matrix. We have
    $$
    operatorname{trace}(M) = begin{cases}
    a_{(n+1)/2} & n text{ is odd}\
    0 & n text{ is even}
    end{cases}
    $$
    and $det(M) = (-1)^{lfloor n/2rfloor} a_1 cdots a_n$, where $lfloor x rfloor$ is the greatest integer $m$ such that $m leq x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why you have used greatest integer function?
      $endgroup$
      – sejy
      Dec 21 '18 at 17:37






    • 1




      $begingroup$
      In order to capture the sign change that the other answer explains more carefully
      $endgroup$
      – Omnomnomnom
      Dec 21 '18 at 17:38











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $n$ is even, then the trace is zero since all diagonal elements are zero. If $n$ is odd, then the $frac{n+1}{2}, frac{n+1}{2}$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.



    The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $-1$ depending on the value of $n$ (specifically, if $n equiv 2 mod 4$ or $n equiv 3 mod 4$ then you need to multiply by $-1$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $n$ is even, then the trace is zero since all diagonal elements are zero. If $n$ is odd, then the $frac{n+1}{2}, frac{n+1}{2}$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.



      The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $-1$ depending on the value of $n$ (specifically, if $n equiv 2 mod 4$ or $n equiv 3 mod 4$ then you need to multiply by $-1$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $n$ is even, then the trace is zero since all diagonal elements are zero. If $n$ is odd, then the $frac{n+1}{2}, frac{n+1}{2}$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.



        The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $-1$ depending on the value of $n$ (specifically, if $n equiv 2 mod 4$ or $n equiv 3 mod 4$ then you need to multiply by $-1$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.






        share|cite|improve this answer









        $endgroup$



        The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $n$ is even, then the trace is zero since all diagonal elements are zero. If $n$ is odd, then the $frac{n+1}{2}, frac{n+1}{2}$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.



        The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $-1$ depending on the value of $n$ (specifically, if $n equiv 2 mod 4$ or $n equiv 3 mod 4$ then you need to multiply by $-1$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 17:32









        angryavianangryavian

        42.4k23481




        42.4k23481























            1












            $begingroup$

            Take
            $$
            M = pmatrix{&&a_1\&cdots\a_n}
            $$
            to be your anti-diagonal matrix. We have
            $$
            operatorname{trace}(M) = begin{cases}
            a_{(n+1)/2} & n text{ is odd}\
            0 & n text{ is even}
            end{cases}
            $$
            and $det(M) = (-1)^{lfloor n/2rfloor} a_1 cdots a_n$, where $lfloor x rfloor$ is the greatest integer $m$ such that $m leq x$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why you have used greatest integer function?
              $endgroup$
              – sejy
              Dec 21 '18 at 17:37






            • 1




              $begingroup$
              In order to capture the sign change that the other answer explains more carefully
              $endgroup$
              – Omnomnomnom
              Dec 21 '18 at 17:38
















            1












            $begingroup$

            Take
            $$
            M = pmatrix{&&a_1\&cdots\a_n}
            $$
            to be your anti-diagonal matrix. We have
            $$
            operatorname{trace}(M) = begin{cases}
            a_{(n+1)/2} & n text{ is odd}\
            0 & n text{ is even}
            end{cases}
            $$
            and $det(M) = (-1)^{lfloor n/2rfloor} a_1 cdots a_n$, where $lfloor x rfloor$ is the greatest integer $m$ such that $m leq x$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why you have used greatest integer function?
              $endgroup$
              – sejy
              Dec 21 '18 at 17:37






            • 1




              $begingroup$
              In order to capture the sign change that the other answer explains more carefully
              $endgroup$
              – Omnomnomnom
              Dec 21 '18 at 17:38














            1












            1








            1





            $begingroup$

            Take
            $$
            M = pmatrix{&&a_1\&cdots\a_n}
            $$
            to be your anti-diagonal matrix. We have
            $$
            operatorname{trace}(M) = begin{cases}
            a_{(n+1)/2} & n text{ is odd}\
            0 & n text{ is even}
            end{cases}
            $$
            and $det(M) = (-1)^{lfloor n/2rfloor} a_1 cdots a_n$, where $lfloor x rfloor$ is the greatest integer $m$ such that $m leq x$.






            share|cite|improve this answer









            $endgroup$



            Take
            $$
            M = pmatrix{&&a_1\&cdots\a_n}
            $$
            to be your anti-diagonal matrix. We have
            $$
            operatorname{trace}(M) = begin{cases}
            a_{(n+1)/2} & n text{ is odd}\
            0 & n text{ is even}
            end{cases}
            $$
            and $det(M) = (-1)^{lfloor n/2rfloor} a_1 cdots a_n$, where $lfloor x rfloor$ is the greatest integer $m$ such that $m leq x$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 21 '18 at 17:32









            OmnomnomnomOmnomnomnom

            129k792185




            129k792185












            • $begingroup$
              why you have used greatest integer function?
              $endgroup$
              – sejy
              Dec 21 '18 at 17:37






            • 1




              $begingroup$
              In order to capture the sign change that the other answer explains more carefully
              $endgroup$
              – Omnomnomnom
              Dec 21 '18 at 17:38


















            • $begingroup$
              why you have used greatest integer function?
              $endgroup$
              – sejy
              Dec 21 '18 at 17:37






            • 1




              $begingroup$
              In order to capture the sign change that the other answer explains more carefully
              $endgroup$
              – Omnomnomnom
              Dec 21 '18 at 17:38
















            $begingroup$
            why you have used greatest integer function?
            $endgroup$
            – sejy
            Dec 21 '18 at 17:37




            $begingroup$
            why you have used greatest integer function?
            $endgroup$
            – sejy
            Dec 21 '18 at 17:37




            1




            1




            $begingroup$
            In order to capture the sign change that the other answer explains more carefully
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:38




            $begingroup$
            In order to capture the sign change that the other answer explains more carefully
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:38


















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