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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045297%2fsimultaneous-diagonalisation-of-quadratic-forms%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045297%2fsimultaneous-diagonalisation-of-quadratic-forms%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                      Read more

                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                      Image
                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                      Read more