How to prove surjectivity of determinant












2












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It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?






determinant







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$endgroup$












  • $begingroup$
    Hint: use a diagonal matrix.
    $endgroup$
    – Bernard
    Jan 8 at 13:11
















2












$begingroup$


It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?






determinant







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: use a diagonal matrix.
    $endgroup$
    – Bernard
    Jan 8 at 13:11














2












2








2





$begingroup$


It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?






determinant







share|cite|improve this question









$endgroup$




It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?






determinant




determinant






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share|cite|improve this question











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asked Jan 8 at 13:09









Zakariae BenslimaneZakariae Benslimane

11010




11010












  • $begingroup$
    Hint: use a diagonal matrix.
    $endgroup$
    – Bernard
    Jan 8 at 13:11


















  • $begingroup$
    Hint: use a diagonal matrix.
    $endgroup$
    – Bernard
    Jan 8 at 13:11
















$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11




$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11










3 Answers
3






active

oldest

votes


















3












$begingroup$

Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.






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  • $begingroup$
    obivous, but took me a while to find the same answer by myself. thank you for the hint
    $endgroup$
    – Zakariae Benslimane
    Jan 8 at 13:43



















3












$begingroup$

Hint:



Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)



    So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        obivous, but took me a while to find the same answer by myself. thank you for the hint
        $endgroup$
        – Zakariae Benslimane
        Jan 8 at 13:43
















      3












      $begingroup$

      Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        obivous, but took me a while to find the same answer by myself. thank you for the hint
        $endgroup$
        – Zakariae Benslimane
        Jan 8 at 13:43














      3












      3








      3





      $begingroup$

      Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.






      share|cite|improve this answer









      $endgroup$



      Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 8 at 13:11









      SamboSambo

      2,3012532




      2,3012532












      • $begingroup$
        obivous, but took me a while to find the same answer by myself. thank you for the hint
        $endgroup$
        – Zakariae Benslimane
        Jan 8 at 13:43


















      • $begingroup$
        obivous, but took me a while to find the same answer by myself. thank you for the hint
        $endgroup$
        – Zakariae Benslimane
        Jan 8 at 13:43
















      $begingroup$
      obivous, but took me a while to find the same answer by myself. thank you for the hint
      $endgroup$
      – Zakariae Benslimane
      Jan 8 at 13:43




      $begingroup$
      obivous, but took me a while to find the same answer by myself. thank you for the hint
      $endgroup$
      – Zakariae Benslimane
      Jan 8 at 13:43











      3












      $begingroup$

      Hint:



      Given $rin Bbb R $, take the diagonal matrix
      $begin{bmatrix}r&0&cdots & 0\
      0&1&cdots & 0\
      vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $      .






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Hint:



        Given $rin Bbb R $, take the diagonal matrix
        $begin{bmatrix}r&0&cdots & 0\
        0&1&cdots & 0\
        vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $        .






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint:



          Given $rin Bbb R $, take the diagonal matrix
          $begin{bmatrix}r&0&cdots & 0\
          0&1&cdots & 0\
          vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $          .






          share|cite|improve this answer











          $endgroup$



          Hint:



          Given $rin Bbb R $, take the diagonal matrix
          $begin{bmatrix}r&0&cdots & 0\
          0&1&cdots & 0\
          vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $          .







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 13:19

























          answered Jan 8 at 13:14









          Thomas ShelbyThomas Shelby

          4,5862727




          4,5862727























              2












              $begingroup$

              Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)



              So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)



                So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)



                  So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.






                  share|cite|improve this answer









                  $endgroup$



                  Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)



                  So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 13:39









                  Chris CusterChris Custer

                  14.2k3827




                  14.2k3827






























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