Are Jordan “Formable” matrices closed under multiplication?











up vote
6
down vote

favorite












After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.



I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.



Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.



Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.










share|cite|improve this question







New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
    – I like Serena
    2 days ago






  • 2




    @IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
    – Travis
    2 days ago






  • 2




    Thank you @Travis. Done.
    – I like Serena
    2 days ago















up vote
6
down vote

favorite












After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.



I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.



Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.



Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.










share|cite|improve this question







New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
    – I like Serena
    2 days ago






  • 2




    @IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
    – Travis
    2 days ago






  • 2




    Thank you @Travis. Done.
    – I like Serena
    2 days ago













up vote
6
down vote

favorite









up vote
6
down vote

favorite











After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.



I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.



Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.



Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.










share|cite|improve this question







New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B in M_n(mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.



I began playing around with the $2 times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.



Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.



Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.







linear-algebra abstract-algebra






share|cite|improve this question







New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Josabanks

434




434




New contributor




Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Josabanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
    – I like Serena
    2 days ago






  • 2




    @IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
    – Travis
    2 days ago






  • 2




    Thank you @Travis. Done.
    – I like Serena
    2 days ago














  • 3




    Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
    – I like Serena
    2 days ago






  • 2




    @IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
    – Travis
    2 days ago






  • 2




    Thank you @Travis. Done.
    – I like Serena
    2 days ago








3




3




Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
2 days ago




Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues.
– I like Serena
2 days ago




2




2




@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
2 days ago




@IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer?
– Travis
2 days ago




2




2




Thank you @Travis. Done.
– I like Serena
2 days ago




Thank you @Travis. Done.
– I like Serena
2 days ago










3 Answers
3






active

oldest

votes

















up vote
6
down vote



accepted










This is false already in the $2 times 2$ case. Consider the matrices



$$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$



$B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But



$$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$



has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.



I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.






share|cite|improve this answer






























    up vote
    7
    down vote













    Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.



    So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.






    share|cite|improve this answer



















    • 1




      Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
      – Daniel Schepler
      2 days ago










    • I've clarified that it's 'generally'.
      – I like Serena
      2 days ago


















    up vote
    3
    down vote













    Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).



    If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.



    It follows immediately that the answer to your question is negative unless $n=1$.






    share|cite|improve this answer























      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',
      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
      });


      }
      });






      Josabanks is a new contributor. Be nice, and check out our Code of Conduct.










       

      draft saved


      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000326%2fare-jordan-formable-matrices-closed-under-multiplication%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      6
      down vote



      accepted










      This is false already in the $2 times 2$ case. Consider the matrices



      $$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$



      $B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But



      $$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$



      has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.



      I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.






      share|cite|improve this answer



























        up vote
        6
        down vote



        accepted










        This is false already in the $2 times 2$ case. Consider the matrices



        $$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$



        $B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But



        $$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$



        has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.



        I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.






        share|cite|improve this answer

























          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          This is false already in the $2 times 2$ case. Consider the matrices



          $$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$



          $B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But



          $$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$



          has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.



          I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.






          share|cite|improve this answer














          This is false already in the $2 times 2$ case. Consider the matrices



          $$A = left[ begin{array}{cc} -1 & 2 \ -2 & 3 end{array} right], B = left[ begin{array}{cc} 3 & 0 \ 0 & 1 end{array} right]$$



          $B$ has eigenvalues $3, 1$, and the characteristic polynomial of $A$ is $t^2 - 2t + 1 = (t - 1)^2$, so $A$ has eigenvalues $1, 1$. But



          $$AB = left[ begin{array}{cc} -3 & 2 \ -6 & 3 end{array} right]$$



          has characteristic polynomial $t^2 + 3$, so has eigenvalues $pm i sqrt{3}$.



          I found these matrices by first taking $B$ to be diagonal with diagonal entries $s, 1$ and then looked at the effect multiplication by $B$ had on the characteristic polynomial of a generic $A$. It's possible to choose $s$ so that $AB$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago









          Trevor Gunn

          13.6k32045




          13.6k32045










          answered 2 days ago









          Qiaochu Yuan

          274k32578914




          274k32578914






















              up vote
              7
              down vote













              Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.



              So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.






              share|cite|improve this answer



















              • 1




                Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
                – Daniel Schepler
                2 days ago










              • I've clarified that it's 'generally'.
                – I like Serena
                2 days ago















              up vote
              7
              down vote













              Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.



              So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.






              share|cite|improve this answer



















              • 1




                Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
                – Daniel Schepler
                2 days ago










              • I've clarified that it's 'generally'.
                – I like Serena
                2 days ago













              up vote
              7
              down vote










              up vote
              7
              down vote









              Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.



              So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.






              share|cite|improve this answer














              Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.



              So the product $AB$ of two real matrices $A$ and $B$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 days ago

























              answered 2 days ago









              I like Serena

              3,1981718




              3,1981718








              • 1




                Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
                – Daniel Schepler
                2 days ago










              • I've clarified that it's 'generally'.
                – I like Serena
                2 days ago














              • 1




                Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
                – Daniel Schepler
                2 days ago










              • I've clarified that it's 'generally'.
                – I like Serena
                2 days ago








              1




              1




              Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
              – Daniel Schepler
              2 days ago




              Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $pi$, or $-I$ which is diagonal.
              – Daniel Schepler
              2 days ago












              I've clarified that it's 'generally'.
              – I like Serena
              2 days ago




              I've clarified that it's 'generally'.
              – I like Serena
              2 days ago










              up vote
              3
              down vote













              Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).



              If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.



              It follows immediately that the answer to your question is negative unless $n=1$.






              share|cite|improve this answer



























                up vote
                3
                down vote













                Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).



                If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.



                It follows immediately that the answer to your question is negative unless $n=1$.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).



                  If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.



                  It follows immediately that the answer to your question is negative unless $n=1$.






                  share|cite|improve this answer














                  Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).



                  If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Consequently, every square matrix is a product of two symmetric matrices (because $A=(SA^T)S^{-1}$) and one of them is non-singular.



                  It follows immediately that the answer to your question is negative unless $n=1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  user1551

                  69.9k566124




                  69.9k566124






















                      Josabanks is a new contributor. Be nice, and check out our Code of Conduct.










                       

                      draft saved


                      draft discarded


















                      Josabanks is a new contributor. Be nice, and check out our Code of Conduct.













                      Josabanks is a new contributor. Be nice, and check out our Code of Conduct.












                      Josabanks is a new contributor. Be nice, and check out our Code of Conduct.















                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000326%2fare-jordan-formable-matrices-closed-under-multiplication%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

                      數位音樂下載

                      When can things happen in Etherscan, such as the picture below?

                      格利澤436b