Calculate minimal polynomial of a matrix












4












$begingroup$


begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






linear-algebra matrices minimal-polynomials







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
    I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






    linear-algebra matrices minimal-polynomials







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      5



      $begingroup$


      begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
      I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






      linear-algebra matrices minimal-polynomials







      share|cite|improve this question











      $endgroup$




      begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
      I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






      linear-algebra matrices minimal-polynomials




      linear-algebra matrices minimal-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 11 '17 at 4:19









      Parcly Taxel

      1




      1










      asked May 11 '17 at 4:08









      user439852user439852

      27113




      27113






















          3 Answers
          3






          active

          oldest

          votes


















          9












          $begingroup$

          These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



          Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
          $$pmatrix{1\0\0\0}overset Amapsto
          pmatrix{0\1\0\1}overset Amapsto
          pmatrix{2\0\2\0}overset Amapsto
          pmatrix{0\4\0\4}
          $$
          with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



          As you see, one can do entirely without the characteristic polynomial.






          share|cite|improve this answer











          $endgroup$





















            7












            $begingroup$

            All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



            Hence $x(x^2-4)$ divides the minimal polynomial,



            Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



            Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



            Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



            Another way to decide on the last part:
            The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
              $endgroup$
              – Pedro Tamaroff♦
              Dec 5 '18 at 15:57



















            1












            $begingroup$

            As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
              $endgroup$
              – Jose Brox
              Dec 5 '18 at 20:08











            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%2f2275809%2fcalculate-minimal-polynomial-of-a-matrix%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









            9












            $begingroup$

            These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



            Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
            $$pmatrix{1\0\0\0}overset Amapsto
            pmatrix{0\1\0\1}overset Amapsto
            pmatrix{2\0\2\0}overset Amapsto
            pmatrix{0\4\0\4}
            $$
            with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



            As you see, one can do entirely without the characteristic polynomial.






            share|cite|improve this answer











            $endgroup$


















              9












              $begingroup$

              These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



              Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
              $$pmatrix{1\0\0\0}overset Amapsto
              pmatrix{0\1\0\1}overset Amapsto
              pmatrix{2\0\2\0}overset Amapsto
              pmatrix{0\4\0\4}
              $$
              with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



              As you see, one can do entirely without the characteristic polynomial.






              share|cite|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



                Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
                $$pmatrix{1\0\0\0}overset Amapsto
                pmatrix{0\1\0\1}overset Amapsto
                pmatrix{2\0\2\0}overset Amapsto
                pmatrix{0\4\0\4}
                $$
                with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



                As you see, one can do entirely without the characteristic polynomial.






                share|cite|improve this answer











                $endgroup$



                These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



                Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
                $$pmatrix{1\0\0\0}overset Amapsto
                pmatrix{0\1\0\1}overset Amapsto
                pmatrix{2\0\2\0}overset Amapsto
                pmatrix{0\4\0\4}
                $$
                with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



                As you see, one can do entirely without the characteristic polynomial.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited May 11 '17 at 9:25

























                answered May 11 '17 at 4:42









                Marc van LeeuwenMarc van Leeuwen

                87.6k5110225




                87.6k5110225























                    7












                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57
















                    7












                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57














                    7












                    7








                    7





                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$



                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited May 11 '17 at 4:43

























                    answered May 11 '17 at 4:33









                    Arpan1729Arpan1729

                    2,7871320




                    2,7871320








                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57














                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57








                    1




                    1




                    $begingroup$
                    Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                    $endgroup$
                    – Pedro Tamaroff♦
                    Dec 5 '18 at 15:57




                    $begingroup$
                    Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                    $endgroup$
                    – Pedro Tamaroff♦
                    Dec 5 '18 at 15:57











                    1












                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08
















                    1












                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08














                    1












                    1








                    1





                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$



                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 19 '18 at 5:22









                    user454200user454200

                    412




                    412












                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08


















                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08
















                    $begingroup$
                    @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                    $endgroup$
                    – Jose Brox
                    Dec 5 '18 at 20:08




                    $begingroup$
                    @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                    $endgroup$
                    – Jose Brox
                    Dec 5 '18 at 20:08


















                    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%2f2275809%2fcalculate-minimal-polynomial-of-a-matrix%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

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                    Read more

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                    Image
                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                    Read more