Question about the Smith normal form of a matrix over the field of integers












1












$begingroup$


I understand that I would need to perform elementary row and column operations. So for a matrix
$M =
begin{array}{cc}
-1 & 1 \
0 & 2 \
end{array}$



am I correct in saying that it cannot be reduced further than $M =
begin{array}{cc}
1 & 0 \
0 & 2 \
end{array}$?

Same thing with any diagonal matrix say $M =
begin{array}{cc}
2 & 0 \
0 & 5 \
end{array}$ to begin with, it cannot be reduced further, right? Appreciate any hint. I was just introduced to normal forms.






matrices







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I understand that I would need to perform elementary row and column operations. So for a matrix
    $M =
    begin{array}{cc}
    -1 & 1 \
    0 & 2 \
    end{array}$



    am I correct in saying that it cannot be reduced further than $M =
    begin{array}{cc}
    1 & 0 \
    0 & 2 \
    end{array}$    ?

    Same thing with any diagonal matrix say $M =
    begin{array}{cc}
    2 & 0 \
    0 & 5 \
    end{array}$     to begin with, it cannot be reduced further, right? Appreciate any hint. I was just introduced to normal forms.






    matrices







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I understand that I would need to perform elementary row and column operations. So for a matrix
      $M =
      begin{array}{cc}
      -1 & 1 \
      0 & 2 \
      end{array}$



      am I correct in saying that it cannot be reduced further than $M =
      begin{array}{cc}
      1 & 0 \
      0 & 2 \
      end{array}$      ?

      Same thing with any diagonal matrix say $M =
      begin{array}{cc}
      2 & 0 \
      0 & 5 \
      end{array}$       to begin with, it cannot be reduced further, right? Appreciate any hint. I was just introduced to normal forms.






      matrices







      share|cite|improve this question











      $endgroup$




      I understand that I would need to perform elementary row and column operations. So for a matrix
      $M =
      begin{array}{cc}
      -1 & 1 \
      0 & 2 \
      end{array}$



      am I correct in saying that it cannot be reduced further than $M =
      begin{array}{cc}
      1 & 0 \
      0 & 2 \
      end{array}$      ?

      Same thing with any diagonal matrix say $M =
      begin{array}{cc}
      2 & 0 \
      0 & 5 \
      end{array}$       to begin with, it cannot be reduced further, right? Appreciate any hint. I was just introduced to normal forms.






      matrices




      matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 7:06







      manifolded

















      asked Jan 30 at 7:03









      manifoldedmanifolded

      49519




      49519






















          1 Answer
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          1












          $begingroup$

          The Smith normal form of
          $pmatrix{2&0\0&5}$ is $pmatrix{1&0\0&10}$. More generally, the SNF of
          $pmatrix{a&0\0&b}$ is $pmatrix{gcd(a,b)&0\0&text{lcm}(a,b)}$.



          With $pmatrix{2&0\0&5}$ you can add column $2$ to column $1$ to get
          $pmatrix{2&0\5&5}$ and then you can reduce the first column via row operations
          to $pmatrix{1\0}$. The second column will then be $pmatrix{*\pm10}$
          and one finally gets to $pmatrix{1&0\0&10}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect!!!!!!!!
            $endgroup$
            – manifolded
            Jan 30 at 7:29












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          1 Answer
          1






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          oldest

          votes









          active

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          active

          oldest

          votes









          1












          $begingroup$

          The Smith normal form of
          $pmatrix{2&0\0&5}$ is $pmatrix{1&0\0&10}$. More generally, the SNF of
          $pmatrix{a&0\0&b}$ is $pmatrix{gcd(a,b)&0\0&text{lcm}(a,b)}$.



          With $pmatrix{2&0\0&5}$ you can add column $2$ to column $1$ to get
          $pmatrix{2&0\5&5}$ and then you can reduce the first column via row operations
          to $pmatrix{1\0}$. The second column will then be $pmatrix{*\pm10}$
          and one finally gets to $pmatrix{1&0\0&10}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect!!!!!!!!
            $endgroup$
            – manifolded
            Jan 30 at 7:29
















          1












          $begingroup$

          The Smith normal form of
          $pmatrix{2&0\0&5}$ is $pmatrix{1&0\0&10}$. More generally, the SNF of
          $pmatrix{a&0\0&b}$ is $pmatrix{gcd(a,b)&0\0&text{lcm}(a,b)}$.



          With $pmatrix{2&0\0&5}$ you can add column $2$ to column $1$ to get
          $pmatrix{2&0\5&5}$ and then you can reduce the first column via row operations
          to $pmatrix{1\0}$. The second column will then be $pmatrix{*\pm10}$
          and one finally gets to $pmatrix{1&0\0&10}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect!!!!!!!!
            $endgroup$
            – manifolded
            Jan 30 at 7:29














          1












          1








          1





          $begingroup$

          The Smith normal form of
          $pmatrix{2&0\0&5}$ is $pmatrix{1&0\0&10}$. More generally, the SNF of
          $pmatrix{a&0\0&b}$ is $pmatrix{gcd(a,b)&0\0&text{lcm}(a,b)}$.



          With $pmatrix{2&0\0&5}$ you can add column $2$ to column $1$ to get
          $pmatrix{2&0\5&5}$ and then you can reduce the first column via row operations
          to $pmatrix{1\0}$. The second column will then be $pmatrix{*\pm10}$
          and one finally gets to $pmatrix{1&0\0&10}$.






          share|cite|improve this answer









          $endgroup$



          The Smith normal form of
          $pmatrix{2&0\0&5}$ is $pmatrix{1&0\0&10}$. More generally, the SNF of
          $pmatrix{a&0\0&b}$ is $pmatrix{gcd(a,b)&0\0&text{lcm}(a,b)}$.



          With $pmatrix{2&0\0&5}$ you can add column $2$ to column $1$ to get
          $pmatrix{2&0\5&5}$ and then you can reduce the first column via row operations
          to $pmatrix{1\0}$. The second column will then be $pmatrix{*\pm10}$
          and one finally gets to $pmatrix{1&0\0&10}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 7:11









          Lord Shark the UnknownLord Shark the Unknown

          108k1162135




          108k1162135












          • $begingroup$
            Perfect!!!!!!!!
            $endgroup$
            – manifolded
            Jan 30 at 7:29


















          • $begingroup$
            Perfect!!!!!!!!
            $endgroup$
            – manifolded
            Jan 30 at 7:29
















          $begingroup$
          Perfect!!!!!!!!
          $endgroup$
          – manifolded
          Jan 30 at 7:29




          $begingroup$
          Perfect!!!!!!!!
          $endgroup$
          – manifolded
          Jan 30 at 7:29


















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