Simultaneous diagonalisation of quadratic forms












0












$begingroup$


Is there a linear transformation that simultaneously reduces the pair of real quadratic forms
$$x^2-y^2$$ and $$2xy$$ to diagonal forms?



My attempt I know that neither of these forms are positive definite, but if
P=$$ left[
begin{array}{ c c }
2 & 0 \
1 & 0
end{array} right]
$$
so that $x=2X$ and $y=X$ then both are diagonal. Is this correct?






linear-algebra matrices diagonalization quadratic-forms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The given matrix $P$ is not invertible and so cannot be used for diagonalization.
    $endgroup$
    – Travis
    Dec 18 '18 at 15:50
















0












$begingroup$


Is there a linear transformation that simultaneously reduces the pair of real quadratic forms
$$x^2-y^2$$ and $$2xy$$ to diagonal forms?



My attempt I know that neither of these forms are positive definite, but if
P=$$ left[
begin{array}{ c c }
2 & 0 \
1 & 0
end{array} right]
$$
so that $x=2X$ and $y=X$ then both are diagonal. Is this correct?






linear-algebra matrices diagonalization quadratic-forms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The given matrix $P$ is not invertible and so cannot be used for diagonalization.
    $endgroup$
    – Travis
    Dec 18 '18 at 15:50














0












0








0





$begingroup$


Is there a linear transformation that simultaneously reduces the pair of real quadratic forms
$$x^2-y^2$$ and $$2xy$$ to diagonal forms?



My attempt I know that neither of these forms are positive definite, but if
P=$$ left[
begin{array}{ c c }
2 & 0 \
1 & 0
end{array} right]
$$
so that $x=2X$ and $y=X$ then both are diagonal. Is this correct?






linear-algebra matrices diagonalization quadratic-forms







share|cite|improve this question











$endgroup$




Is there a linear transformation that simultaneously reduces the pair of real quadratic forms
$$x^2-y^2$$ and $$2xy$$ to diagonal forms?



My attempt I know that neither of these forms are positive definite, but if
P=$$ left[
begin{array}{ c c }
2 & 0 \
1 & 0
end{array} right]
$$
so that $x=2X$ and $y=X$ then both are diagonal. Is this correct?






linear-algebra matrices diagonalization quadratic-forms




linear-algebra matrices diagonalization quadratic-forms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 16:33









Travis

63.8k769151




63.8k769151










asked Dec 18 '18 at 15:35









MathematicianPMathematicianP

3416




3416








  • 1




    $begingroup$
    The given matrix $P$ is not invertible and so cannot be used for diagonalization.
    $endgroup$
    – Travis
    Dec 18 '18 at 15:50














  • 1




    $begingroup$
    The given matrix $P$ is not invertible and so cannot be used for diagonalization.
    $endgroup$
    – Travis
    Dec 18 '18 at 15:50








1




1




$begingroup$
The given matrix $P$ is not invertible and so cannot be used for diagonalization.
$endgroup$
– Travis
Dec 18 '18 at 15:50




$begingroup$
The given matrix $P$ is not invertible and so cannot be used for diagonalization.
$endgroup$
– Travis
Dec 18 '18 at 15:50










2 Answers
2






active

oldest

votes


















1












$begingroup$

First note that the given matrix $P$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.



The matrix representations of the two quadratic forms with respect to the standard basis are
$$Q = pmatrix{1&\&-1}, qquad Q' = pmatrix{&1\1&}.$$



Computing for a general change-of-basis matrix $P = pmatrix{p_{ij}}$ the matrix representations $P^{top} Q P, P^{top} Q' P$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components
$$a = p_{11} p_{12} - p_{21} p_{22}, qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$
and by definition an invertible matrix $P$ simultaneously diagonalizes $Q, Q'$ iff $a = b = 0$.




For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $P$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $Bbb R$. On the other hand, over $Bbb C$ this equation doesn't force degeneracy of $P$, only that $p_{21} = pm i p_{11}$ or $p_{12} = pm i p_{21}$. Substituting quickly leads to the solutions $P = pmatrix{lambda&pm imu\pm ilambda&mu}$, so the quadratic forms are simultaneously diagonalizable over $Bbb C$.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You need an invertible matrix for diagonalization, so your $P$ doesn’t work. It collapses everything onto a single line.



    You can approach this problem geometrically. Restricted to the reals, $x^2-y^2=a$ is a family of hyperbolas with common asymptotes $x=y$ and $x=-y$, and $2xy=b$ is a family of hyperbolas with common asmyptotes $x=0$ and $y=0$. Simultaneous diagonalization makes these asymptotes coincide, but $x=0$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      1












      $begingroup$

      First note that the given matrix $P$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.



      The matrix representations of the two quadratic forms with respect to the standard basis are
      $$Q = pmatrix{1&\&-1}, qquad Q' = pmatrix{&1\1&}.$$



      Computing for a general change-of-basis matrix $P = pmatrix{p_{ij}}$ the matrix representations $P^{top} Q P, P^{top} Q' P$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components
      $$a = p_{11} p_{12} - p_{21} p_{22}, qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$
      and by definition an invertible matrix $P$ simultaneously diagonalizes $Q, Q'$ iff $a = b = 0$.




      For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $P$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $Bbb R$. On the other hand, over $Bbb C$ this equation doesn't force degeneracy of $P$, only that $p_{21} = pm i p_{11}$ or $p_{12} = pm i p_{21}$. Substituting quickly leads to the solutions $P = pmatrix{lambda&pm imu\pm ilambda&mu}$, so the quadratic forms are simultaneously diagonalizable over $Bbb C$.







      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        First note that the given matrix $P$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.



        The matrix representations of the two quadratic forms with respect to the standard basis are
        $$Q = pmatrix{1&\&-1}, qquad Q' = pmatrix{&1\1&}.$$



        Computing for a general change-of-basis matrix $P = pmatrix{p_{ij}}$ the matrix representations $P^{top} Q P, P^{top} Q' P$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components
        $$a = p_{11} p_{12} - p_{21} p_{22}, qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$
        and by definition an invertible matrix $P$ simultaneously diagonalizes $Q, Q'$ iff $a = b = 0$.




        For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $P$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $Bbb R$. On the other hand, over $Bbb C$ this equation doesn't force degeneracy of $P$, only that $p_{21} = pm i p_{11}$ or $p_{12} = pm i p_{21}$. Substituting quickly leads to the solutions $P = pmatrix{lambda&pm imu\pm ilambda&mu}$, so the quadratic forms are simultaneously diagonalizable over $Bbb C$.







        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          First note that the given matrix $P$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.



          The matrix representations of the two quadratic forms with respect to the standard basis are
          $$Q = pmatrix{1&\&-1}, qquad Q' = pmatrix{&1\1&}.$$



          Computing for a general change-of-basis matrix $P = pmatrix{p_{ij}}$ the matrix representations $P^{top} Q P, P^{top} Q' P$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components
          $$a = p_{11} p_{12} - p_{21} p_{22}, qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$
          and by definition an invertible matrix $P$ simultaneously diagonalizes $Q, Q'$ iff $a = b = 0$.




          For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $P$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $Bbb R$. On the other hand, over $Bbb C$ this equation doesn't force degeneracy of $P$, only that $p_{21} = pm i p_{11}$ or $p_{12} = pm i p_{21}$. Substituting quickly leads to the solutions $P = pmatrix{lambda&pm imu\pm ilambda&mu}$, so the quadratic forms are simultaneously diagonalizable over $Bbb C$.







          share|cite|improve this answer









          $endgroup$



          First note that the given matrix $P$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.



          The matrix representations of the two quadratic forms with respect to the standard basis are
          $$Q = pmatrix{1&\&-1}, qquad Q' = pmatrix{&1\1&}.$$



          Computing for a general change-of-basis matrix $P = pmatrix{p_{ij}}$ the matrix representations $P^{top} Q P, P^{top} Q' P$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components
          $$a = p_{11} p_{12} - p_{21} p_{22}, qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$
          and by definition an invertible matrix $P$ simultaneously diagonalizes $Q, Q'$ iff $a = b = 0$.




          For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $P$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $Bbb R$. On the other hand, over $Bbb C$ this equation doesn't force degeneracy of $P$, only that $p_{21} = pm i p_{11}$ or $p_{12} = pm i p_{21}$. Substituting quickly leads to the solutions $P = pmatrix{lambda&pm imu\pm ilambda&mu}$, so the quadratic forms are simultaneously diagonalizable over $Bbb C$.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 16:32









          TravisTravis

          63.8k769151




          63.8k769151























              0












              $begingroup$

              You need an invertible matrix for diagonalization, so your $P$ doesn’t work. It collapses everything onto a single line.



              You can approach this problem geometrically. Restricted to the reals, $x^2-y^2=a$ is a family of hyperbolas with common asymptotes $x=y$ and $x=-y$, and $2xy=b$ is a family of hyperbolas with common asmyptotes $x=0$ and $y=0$. Simultaneous diagonalization makes these asymptotes coincide, but $x=0$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You need an invertible matrix for diagonalization, so your $P$ doesn’t work. It collapses everything onto a single line.



                You can approach this problem geometrically. Restricted to the reals, $x^2-y^2=a$ is a family of hyperbolas with common asymptotes $x=y$ and $x=-y$, and $2xy=b$ is a family of hyperbolas with common asmyptotes $x=0$ and $y=0$. Simultaneous diagonalization makes these asymptotes coincide, but $x=0$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You need an invertible matrix for diagonalization, so your $P$ doesn’t work. It collapses everything onto a single line.



                  You can approach this problem geometrically. Restricted to the reals, $x^2-y^2=a$ is a family of hyperbolas with common asymptotes $x=y$ and $x=-y$, and $2xy=b$ is a family of hyperbolas with common asmyptotes $x=0$ and $y=0$. Simultaneous diagonalization makes these asymptotes coincide, but $x=0$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.






                  share|cite|improve this answer









                  $endgroup$



                  You need an invertible matrix for diagonalization, so your $P$ doesn’t work. It collapses everything onto a single line.



                  You can approach this problem geometrically. Restricted to the reals, $x^2-y^2=a$ is a family of hyperbolas with common asymptotes $x=y$ and $x=-y$, and $2xy=b$ is a family of hyperbolas with common asmyptotes $x=0$ and $y=0$. Simultaneous diagonalization makes these asymptotes coincide, but $x=0$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 20:25









                  amdamd

                  31.3k21051




                  31.3k21051






























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