Is the transformation matrix of an upper triangular matrix to its Jordan normal form always triangular?












1












$begingroup$


Assume that $A$ is an upper triangular matrix. In the case where $A$ is 2x2, I've checked that a transformation matrix $P$ such that $J = P^{-1}AP$, with $J$ Jordan normal form of $A$, is always upper triangular. I think the same is true for every matrix of dimension $n$ (probably a proof by induction), but I've not seen that anywhere. Any source ?






matrices matrix-decomposition jordan-normal-form







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Assume that $A$ is an upper triangular matrix. In the case where $A$ is 2x2, I've checked that a transformation matrix $P$ such that $J = P^{-1}AP$, with $J$ Jordan normal form of $A$, is always upper triangular. I think the same is true for every matrix of dimension $n$ (probably a proof by induction), but I've not seen that anywhere. Any source ?






    matrices matrix-decomposition jordan-normal-form







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Assume that $A$ is an upper triangular matrix. In the case where $A$ is 2x2, I've checked that a transformation matrix $P$ such that $J = P^{-1}AP$, with $J$ Jordan normal form of $A$, is always upper triangular. I think the same is true for every matrix of dimension $n$ (probably a proof by induction), but I've not seen that anywhere. Any source ?






      matrices matrix-decomposition jordan-normal-form







      share|cite|improve this question









      $endgroup$




      Assume that $A$ is an upper triangular matrix. In the case where $A$ is 2x2, I've checked that a transformation matrix $P$ such that $J = P^{-1}AP$, with $J$ Jordan normal form of $A$, is always upper triangular. I think the same is true for every matrix of dimension $n$ (probably a proof by induction), but I've not seen that anywhere. Any source ?






      matrices matrix-decomposition jordan-normal-form




      matrices matrix-decomposition jordan-normal-form






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 12:36









      MikeTeXMikeTeX

      1,278412




      1,278412






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.



          Anyway, this is false for $n = 3$. Indeed, take
          begin{equation}
          A = begin{pmatrix} 2 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 2 end{pmatrix} .
          end{equation}
          Its Jordan normal form is
          begin{equation}
          begin{pmatrix} 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 end{pmatrix} .
          end{equation}
          Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
            $endgroup$
            – MikeTeX
            Jan 10 at 13:04












          • $begingroup$
            @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
            $endgroup$
            – darij grinberg
            Jan 10 at 13:09













          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%2f3068594%2fis-the-transformation-matrix-of-an-upper-triangular-matrix-to-its-jordan-normal%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.



          Anyway, this is false for $n = 3$. Indeed, take
          begin{equation}
          A = begin{pmatrix} 2 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 2 end{pmatrix} .
          end{equation}
          Its Jordan normal form is
          begin{equation}
          begin{pmatrix} 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 end{pmatrix} .
          end{equation}
          Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
            $endgroup$
            – MikeTeX
            Jan 10 at 13:04












          • $begingroup$
            @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
            $endgroup$
            – darij grinberg
            Jan 10 at 13:09


















          1












          $begingroup$

          Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.



          Anyway, this is false for $n = 3$. Indeed, take
          begin{equation}
          A = begin{pmatrix} 2 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 2 end{pmatrix} .
          end{equation}
          Its Jordan normal form is
          begin{equation}
          begin{pmatrix} 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 end{pmatrix} .
          end{equation}
          Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
            $endgroup$
            – MikeTeX
            Jan 10 at 13:04












          • $begingroup$
            @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
            $endgroup$
            – darij grinberg
            Jan 10 at 13:09
















          1












          1








          1





          $begingroup$

          Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.



          Anyway, this is false for $n = 3$. Indeed, take
          begin{equation}
          A = begin{pmatrix} 2 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 2 end{pmatrix} .
          end{equation}
          Its Jordan normal form is
          begin{equation}
          begin{pmatrix} 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 end{pmatrix} .
          end{equation}
          Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.






          share|cite|improve this answer









          $endgroup$



          Pedantic note: You don't want to say that every matrix $P$ satisfying $J = P^{-1} A P$ will be upper-triangular (this is false already for $n = 2$). You want to say that there exists some invertible upper-triangular matrix $P$ satisfying $J = P^{-1} A P$.



          Anyway, this is false for $n = 3$. Indeed, take
          begin{equation}
          A = begin{pmatrix} 2 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 2 end{pmatrix} .
          end{equation}
          Its Jordan normal form is
          begin{equation}
          begin{pmatrix} 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 end{pmatrix} .
          end{equation}
          Thus, in order to bring it into its Jordan normal form, the two $2$'s on its diagonal need to be brought together. But conjugation by an upper-triangular matrix cannot do this: Indeed, it is easy to check that if you conjugate an upper-triangular matrix $B$ by an upper-triangular matrix, then the diagonal entries of $B$ remain unchanged.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 12:52









          darij grinbergdarij grinberg

          11k33167




          11k33167












          • $begingroup$
            It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
            $endgroup$
            – MikeTeX
            Jan 10 at 13:04












          • $begingroup$
            @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
            $endgroup$
            – darij grinberg
            Jan 10 at 13:09




















          • $begingroup$
            It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
            $endgroup$
            – MikeTeX
            Jan 10 at 13:04












          • $begingroup$
            @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
            $endgroup$
            – darij grinberg
            Jan 10 at 13:09


















          $begingroup$
          It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
          $endgroup$
          – MikeTeX
          Jan 10 at 13:04






          $begingroup$
          It would be interesting to find the additional assumptions that make this claim work, maybe, for example, "all the diagonal coefficients are equal".
          $endgroup$
          – MikeTeX
          Jan 10 at 13:04














          $begingroup$
          @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
          $endgroup$
          – darij grinberg
          Jan 10 at 13:09






          $begingroup$
          @MikeTeX: Equality of diagonal entries is not enough. $A = begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix}$ is a counterexample. This time I don't have a slick proof; I just computed the result of conjugating $A$ by a general upper-triangular matrix: $begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} ^{-1}begin{pmatrix} 1 & 0 & 1\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} begin{pmatrix} a & x & y\ 0 & b & z\ 0 & 0 & c end{pmatrix} = begin{pmatrix} 1 & 0 & dfrac{c}{a}\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix} $, which is not a Jordan normal form.
          $endgroup$
          – darij grinberg
          Jan 10 at 13:09




















          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%2f3068594%2fis-the-transformation-matrix-of-an-upper-triangular-matrix-to-its-jordan-normal%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

          Questions related to Moebius Transform of Characteristic Function of the Primes

          Image
          2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
          Read more

          List of scandals in India

          Image
          List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
          Read more

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
          Read more