Jordan form of an $n times n$ Jordan block with eigenvalue $0$












1












$begingroup$


Suppose $J$ is an $n times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?



My solution:



I squared the matrix, it follows that the eigenvalues are $0$ again, and $dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.



Therefore, the Jordan form of $J^2$ is an $(n-2) times (n-2)$ block and two $1times1$ blocks all with eigenvalue zero.



However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.






linear-algebra jordan-normal-form







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Suppose $J$ is an $n times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?



    My solution:



    I squared the matrix, it follows that the eigenvalues are $0$ again, and $dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.



    Therefore, the Jordan form of $J^2$ is an $(n-2) times (n-2)$ block and two $1times1$ blocks all with eigenvalue zero.



    However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.






    linear-algebra jordan-normal-form







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $J$ is an $n times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?



      My solution:



      I squared the matrix, it follows that the eigenvalues are $0$ again, and $dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.



      Therefore, the Jordan form of $J^2$ is an $(n-2) times (n-2)$ block and two $1times1$ blocks all with eigenvalue zero.



      However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.






      linear-algebra jordan-normal-form







      share|cite|improve this question











      $endgroup$




      Suppose $J$ is an $n times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?



      My solution:



      I squared the matrix, it follows that the eigenvalues are $0$ again, and $dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.



      Therefore, the Jordan form of $J^2$ is an $(n-2) times (n-2)$ block and two $1times1$ blocks all with eigenvalue zero.



      However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.






      linear-algebra jordan-normal-form




      linear-algebra jordan-normal-form






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 0:58









      Rócherz

      2,9112821




      2,9112821










      asked Dec 25 '18 at 21:07









      Scosh_lrScosh_lr

      566




      566






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          You're wrong. For example,
          $$ pmatrix{0 & 1 & 0 cr
          0 & 0 & 1cr
          0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1cr 0 & 0 & 0cr 0 & 0 & 0cr}
          text{has Jordan form} pmatrix{0 & 1 & 0cr 0 & 0 & 0cr 0 & 0 & 0cr}$$
          (a block of size $2$ and a block of size $1$) but
          $$ pmatrix{0 & 1 & 0 & 0cr
          0 & 0 & 1 & 0cr
          0 & 0 & 0 & 1cr
          0 & 0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1 & 0cr 0 & 0 & 0 & 1cr
          0 & 0 & 0 & 0cr 0 & 0 & 0 & 0cr} text{has Jordan form}
          pmatrix{0 & 1 & 0 & 0cr 0 & 0 & 0 & 0cr 0 & 0 & 0 & 1cr 0 & 0 & 0 & 0cr}$$
          (two blocks of size $2$).






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let me illustrate this with the $n = 6$ case. Here, we have a basis ${ e_1, e_2, e_3, e_4, e_5, e_6 }$, and the action of $J$ on the basis vectors is given by
            $$ Je_1 = e_2, Je_2 = e_3, Je_3 = e_4, Je_4 = e_5, Je_5 = e_6, Je_6 = 0. $$
            So the action of $J^2$ on this basis is given by



            $$ J^2e_1 = e_3, J^2e_2 = e_4, J^2e_3 = e_5, J^2e_4 = e_6, J^2e_5 = 0, J^2e_6 = 0.$$



            [You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]



            But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:




            • There is the subspace spanned by ${ e_1, e_3, e_5 }$, on which the action of $J^2$ is
              $$ J^2e_1 = e_3, J^2e_3 = e_5, J^2e_5 = 0.$$

            • Then there is the subspace spanned by ${ e_2, e_4, e_6 }$, on which the action of $J^2$ is
              $$ J^2e_2 = e_4, J^2e_4 = e_6, J^2e_6 = 0 .$$


            So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is
            $$ begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{bmatrix}$$



            That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
              $endgroup$
              – Scosh_lr
              Dec 25 '18 at 22:32






            • 1




              $begingroup$
              @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
              $endgroup$
              – Kenny Wong
              Dec 25 '18 at 22:43












            • $begingroup$
              why is it not possible to have a 2x2 block and two 1x1 blocks?
              $endgroup$
              – Scosh_lr
              Dec 25 '18 at 22:57






            • 1




              $begingroup$
              @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
              $endgroup$
              – Kenny Wong
              Dec 25 '18 at 23:24










            • $begingroup$
              Oh I see now, Thanks for the help.
              $endgroup$
              – Scosh_lr
              Dec 25 '18 at 23:26











            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%2f3052424%2fjordan-form-of-an-n-times-n-jordan-block-with-eigenvalue-0%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












            $begingroup$

            You're wrong. For example,
            $$ pmatrix{0 & 1 & 0 cr
            0 & 0 & 1cr
            0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1cr 0 & 0 & 0cr 0 & 0 & 0cr}
            text{has Jordan form} pmatrix{0 & 1 & 0cr 0 & 0 & 0cr 0 & 0 & 0cr}$$
            (a block of size $2$ and a block of size $1$) but
            $$ pmatrix{0 & 1 & 0 & 0cr
            0 & 0 & 1 & 0cr
            0 & 0 & 0 & 1cr
            0 & 0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1 & 0cr 0 & 0 & 0 & 1cr
            0 & 0 & 0 & 0cr 0 & 0 & 0 & 0cr} text{has Jordan form}
            pmatrix{0 & 1 & 0 & 0cr 0 & 0 & 0 & 0cr 0 & 0 & 0 & 1cr 0 & 0 & 0 & 0cr}$$
            (two blocks of size $2$).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You're wrong. For example,
              $$ pmatrix{0 & 1 & 0 cr
              0 & 0 & 1cr
              0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1cr 0 & 0 & 0cr 0 & 0 & 0cr}
              text{has Jordan form} pmatrix{0 & 1 & 0cr 0 & 0 & 0cr 0 & 0 & 0cr}$$
              (a block of size $2$ and a block of size $1$) but
              $$ pmatrix{0 & 1 & 0 & 0cr
              0 & 0 & 1 & 0cr
              0 & 0 & 0 & 1cr
              0 & 0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1 & 0cr 0 & 0 & 0 & 1cr
              0 & 0 & 0 & 0cr 0 & 0 & 0 & 0cr} text{has Jordan form}
              pmatrix{0 & 1 & 0 & 0cr 0 & 0 & 0 & 0cr 0 & 0 & 0 & 1cr 0 & 0 & 0 & 0cr}$$
              (two blocks of size $2$).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You're wrong. For example,
                $$ pmatrix{0 & 1 & 0 cr
                0 & 0 & 1cr
                0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1cr 0 & 0 & 0cr 0 & 0 & 0cr}
                text{has Jordan form} pmatrix{0 & 1 & 0cr 0 & 0 & 0cr 0 & 0 & 0cr}$$
                (a block of size $2$ and a block of size $1$) but
                $$ pmatrix{0 & 1 & 0 & 0cr
                0 & 0 & 1 & 0cr
                0 & 0 & 0 & 1cr
                0 & 0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1 & 0cr 0 & 0 & 0 & 1cr
                0 & 0 & 0 & 0cr 0 & 0 & 0 & 0cr} text{has Jordan form}
                pmatrix{0 & 1 & 0 & 0cr 0 & 0 & 0 & 0cr 0 & 0 & 0 & 1cr 0 & 0 & 0 & 0cr}$$
                (two blocks of size $2$).






                share|cite|improve this answer









                $endgroup$



                You're wrong. For example,
                $$ pmatrix{0 & 1 & 0 cr
                0 & 0 & 1cr
                0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1cr 0 & 0 & 0cr 0 & 0 & 0cr}
                text{has Jordan form} pmatrix{0 & 1 & 0cr 0 & 0 & 0cr 0 & 0 & 0cr}$$
                (a block of size $2$ and a block of size $1$) but
                $$ pmatrix{0 & 1 & 0 & 0cr
                0 & 0 & 1 & 0cr
                0 & 0 & 0 & 1cr
                0 & 0 & 0 & 0cr}^2 = pmatrix{0 & 0 & 1 & 0cr 0 & 0 & 0 & 1cr
                0 & 0 & 0 & 0cr 0 & 0 & 0 & 0cr} text{has Jordan form}
                pmatrix{0 & 1 & 0 & 0cr 0 & 0 & 0 & 0cr 0 & 0 & 0 & 1cr 0 & 0 & 0 & 0cr}$$
                (two blocks of size $2$).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 25 '18 at 21:25









                Robert IsraelRobert Israel

                327k23216469




                327k23216469























                    1












                    $begingroup$

                    Let me illustrate this with the $n = 6$ case. Here, we have a basis ${ e_1, e_2, e_3, e_4, e_5, e_6 }$, and the action of $J$ on the basis vectors is given by
                    $$ Je_1 = e_2, Je_2 = e_3, Je_3 = e_4, Je_4 = e_5, Je_5 = e_6, Je_6 = 0. $$
                    So the action of $J^2$ on this basis is given by



                    $$ J^2e_1 = e_3, J^2e_2 = e_4, J^2e_3 = e_5, J^2e_4 = e_6, J^2e_5 = 0, J^2e_6 = 0.$$



                    [You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]



                    But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:




                    • There is the subspace spanned by ${ e_1, e_3, e_5 }$, on which the action of $J^2$ is
                      $$ J^2e_1 = e_3, J^2e_3 = e_5, J^2e_5 = 0.$$

                    • Then there is the subspace spanned by ${ e_2, e_4, e_6 }$, on which the action of $J^2$ is
                      $$ J^2e_2 = e_4, J^2e_4 = e_6, J^2e_6 = 0 .$$


                    So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is
                    $$ begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{bmatrix}$$



                    That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:32






                    • 1




                      $begingroup$
                      @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 22:43












                    • $begingroup$
                      why is it not possible to have a 2x2 block and two 1x1 blocks?
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:57






                    • 1




                      $begingroup$
                      @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 23:24










                    • $begingroup$
                      Oh I see now, Thanks for the help.
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 23:26
















                    1












                    $begingroup$

                    Let me illustrate this with the $n = 6$ case. Here, we have a basis ${ e_1, e_2, e_3, e_4, e_5, e_6 }$, and the action of $J$ on the basis vectors is given by
                    $$ Je_1 = e_2, Je_2 = e_3, Je_3 = e_4, Je_4 = e_5, Je_5 = e_6, Je_6 = 0. $$
                    So the action of $J^2$ on this basis is given by



                    $$ J^2e_1 = e_3, J^2e_2 = e_4, J^2e_3 = e_5, J^2e_4 = e_6, J^2e_5 = 0, J^2e_6 = 0.$$



                    [You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]



                    But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:




                    • There is the subspace spanned by ${ e_1, e_3, e_5 }$, on which the action of $J^2$ is
                      $$ J^2e_1 = e_3, J^2e_3 = e_5, J^2e_5 = 0.$$

                    • Then there is the subspace spanned by ${ e_2, e_4, e_6 }$, on which the action of $J^2$ is
                      $$ J^2e_2 = e_4, J^2e_4 = e_6, J^2e_6 = 0 .$$


                    So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is
                    $$ begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{bmatrix}$$



                    That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:32






                    • 1




                      $begingroup$
                      @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 22:43












                    • $begingroup$
                      why is it not possible to have a 2x2 block and two 1x1 blocks?
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:57






                    • 1




                      $begingroup$
                      @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 23:24










                    • $begingroup$
                      Oh I see now, Thanks for the help.
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 23:26














                    1












                    1








                    1





                    $begingroup$

                    Let me illustrate this with the $n = 6$ case. Here, we have a basis ${ e_1, e_2, e_3, e_4, e_5, e_6 }$, and the action of $J$ on the basis vectors is given by
                    $$ Je_1 = e_2, Je_2 = e_3, Je_3 = e_4, Je_4 = e_5, Je_5 = e_6, Je_6 = 0. $$
                    So the action of $J^2$ on this basis is given by



                    $$ J^2e_1 = e_3, J^2e_2 = e_4, J^2e_3 = e_5, J^2e_4 = e_6, J^2e_5 = 0, J^2e_6 = 0.$$



                    [You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]



                    But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:




                    • There is the subspace spanned by ${ e_1, e_3, e_5 }$, on which the action of $J^2$ is
                      $$ J^2e_1 = e_3, J^2e_3 = e_5, J^2e_5 = 0.$$

                    • Then there is the subspace spanned by ${ e_2, e_4, e_6 }$, on which the action of $J^2$ is
                      $$ J^2e_2 = e_4, J^2e_4 = e_6, J^2e_6 = 0 .$$


                    So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is
                    $$ begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{bmatrix}$$



                    That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?






                    share|cite|improve this answer









                    $endgroup$



                    Let me illustrate this with the $n = 6$ case. Here, we have a basis ${ e_1, e_2, e_3, e_4, e_5, e_6 }$, and the action of $J$ on the basis vectors is given by
                    $$ Je_1 = e_2, Je_2 = e_3, Je_3 = e_4, Je_4 = e_5, Je_5 = e_6, Je_6 = 0. $$
                    So the action of $J^2$ on this basis is given by



                    $$ J^2e_1 = e_3, J^2e_2 = e_4, J^2e_3 = e_5, J^2e_4 = e_6, J^2e_5 = 0, J^2e_6 = 0.$$



                    [You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]



                    But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:




                    • There is the subspace spanned by ${ e_1, e_3, e_5 }$, on which the action of $J^2$ is
                      $$ J^2e_1 = e_3, J^2e_3 = e_5, J^2e_5 = 0.$$

                    • Then there is the subspace spanned by ${ e_2, e_4, e_6 }$, on which the action of $J^2$ is
                      $$ J^2e_2 = e_4, J^2e_4 = e_6, J^2e_6 = 0 .$$


                    So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is
                    $$ begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{bmatrix}$$



                    That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 25 '18 at 21:26









                    Kenny WongKenny Wong

                    19.1k21441




                    19.1k21441












                    • $begingroup$
                      Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:32






                    • 1




                      $begingroup$
                      @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 22:43












                    • $begingroup$
                      why is it not possible to have a 2x2 block and two 1x1 blocks?
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:57






                    • 1




                      $begingroup$
                      @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 23:24










                    • $begingroup$
                      Oh I see now, Thanks for the help.
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 23:26


















                    • $begingroup$
                      Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:32






                    • 1




                      $begingroup$
                      @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 22:43












                    • $begingroup$
                      why is it not possible to have a 2x2 block and two 1x1 blocks?
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 22:57






                    • 1




                      $begingroup$
                      @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                      $endgroup$
                      – Kenny Wong
                      Dec 25 '18 at 23:24










                    • $begingroup$
                      Oh I see now, Thanks for the help.
                      $endgroup$
                      – Scosh_lr
                      Dec 25 '18 at 23:26
















                    $begingroup$
                    Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 22:32




                    $begingroup$
                    Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 22:32




                    1




                    1




                    $begingroup$
                    @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                    $endgroup$
                    – Kenny Wong
                    Dec 25 '18 at 22:43






                    $begingroup$
                    @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks.
                    $endgroup$
                    – Kenny Wong
                    Dec 25 '18 at 22:43














                    $begingroup$
                    why is it not possible to have a 2x2 block and two 1x1 blocks?
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 22:57




                    $begingroup$
                    why is it not possible to have a 2x2 block and two 1x1 blocks?
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 22:57




                    1




                    1




                    $begingroup$
                    @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                    $endgroup$
                    – Kenny Wong
                    Dec 25 '18 at 23:24




                    $begingroup$
                    @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2times 2$ block, and $e_3$ spans a $1 times 1 $ block and $e_4$ spans the other $1 times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.)
                    $endgroup$
                    – Kenny Wong
                    Dec 25 '18 at 23:24












                    $begingroup$
                    Oh I see now, Thanks for the help.
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 23:26




                    $begingroup$
                    Oh I see now, Thanks for the help.
                    $endgroup$
                    – Scosh_lr
                    Dec 25 '18 at 23:26


















                    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%2f3052424%2fjordan-form-of-an-n-times-n-jordan-block-with-eigenvalue-0%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

                    Wiesbaden

                    Image
                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                    Read more

                    Marschland

                    Image
                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more