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













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      edited Dec 26 '18 at 0:58









      Rócherz

      2,7762721




      2,7762721










      asked Dec 25 '18 at 21:07









      Scosh_lrScosh_lr

      516




      516






















          2 Answers
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          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











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            2 Answers
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            2 Answers
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            active

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            active

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            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

                323k23212466




                323k23212466























                    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

                    18.8k21439




                    18.8k21439












                    • $begingroup$
                      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


















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