Isotropic points with a jacobian matrix












0












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As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:



a   b
c d


Matrix example:



2.x     0
2.y+3 3


How can I find its isotropic points ?

Thank you very much.










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    0












    $begingroup$


    As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:



    a   b
    c d


    Matrix example:



    2.x     0
    2.y+3 3


    How can I find its isotropic points ?

    Thank you very much.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:



      a   b
      c d


      Matrix example:



      2.x     0
      2.y+3 3


      How can I find its isotropic points ?

      Thank you very much.










      share|cite|improve this question









      $endgroup$




      As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:



      a   b
      c d


      Matrix example:



      2.x     0
      2.y+3 3


      How can I find its isotropic points ?

      Thank you very much.







      matrix-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 '14 at 14:31









      XitrumXitrum

      1012




      1012






















          1 Answer
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          $begingroup$

          The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
          $$
          begin{pmatrix}
          a & b \
          c & d
          end{pmatrix}
          $$
          these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.



          For the concrete example
          $$
          begin{pmatrix}
          2x & 0 \
          2y+3 & 3
          end{pmatrix}
          $$
          the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.






          share|cite|improve this answer









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            1 Answer
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            active

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            1












            $begingroup$

            The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
            $$
            begin{pmatrix}
            a & b \
            c & d
            end{pmatrix}
            $$
            these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.



            For the concrete example
            $$
            begin{pmatrix}
            2x & 0 \
            2y+3 & 3
            end{pmatrix}
            $$
            the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
              $$
              begin{pmatrix}
              a & b \
              c & d
              end{pmatrix}
              $$
              these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.



              For the concrete example
              $$
              begin{pmatrix}
              2x & 0 \
              2y+3 & 3
              end{pmatrix}
              $$
              the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
                $$
                begin{pmatrix}
                a & b \
                c & d
                end{pmatrix}
                $$
                these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.



                For the concrete example
                $$
                begin{pmatrix}
                2x & 0 \
                2y+3 & 3
                end{pmatrix}
                $$
                the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.






                share|cite|improve this answer









                $endgroup$



                The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
                $$
                begin{pmatrix}
                a & b \
                c & d
                end{pmatrix}
                $$
                these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.



                For the concrete example
                $$
                begin{pmatrix}
                2x & 0 \
                2y+3 & 3
                end{pmatrix}
                $$
                the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 2 '14 at 0:22









                user127096user127096

                8,17011140




                8,17011140






























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