Minimal polynomial for any power of Jordan block is same as the minimal polynomial of the Jordan block.












1












$begingroup$


Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?



Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.






linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is true for any non-zero eigenvalue, not just eigenvalue $1$.
    $endgroup$
    – user593746
    Dec 6 '18 at 11:19






  • 2




    $begingroup$
    Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
    $endgroup$
    – Eric
    Dec 6 '18 at 15:40








  • 1




    $begingroup$
    The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
    $endgroup$
    – ancientmathematician
    Dec 6 '18 at 16:34










  • $begingroup$
    My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
    $endgroup$
    – user371231
    Dec 6 '18 at 18:46
















1












$begingroup$


Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?



Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.






linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is true for any non-zero eigenvalue, not just eigenvalue $1$.
    $endgroup$
    – user593746
    Dec 6 '18 at 11:19






  • 2




    $begingroup$
    Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
    $endgroup$
    – Eric
    Dec 6 '18 at 15:40








  • 1




    $begingroup$
    The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
    $endgroup$
    – ancientmathematician
    Dec 6 '18 at 16:34










  • $begingroup$
    My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
    $endgroup$
    – user371231
    Dec 6 '18 at 18:46














1












1








1


2



$begingroup$


Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?



Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.






linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation







share|cite|improve this question









$endgroup$




Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?



Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.






linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation




linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 10:12









user371231user371231

756511




756511












  • $begingroup$
    This is true for any non-zero eigenvalue, not just eigenvalue $1$.
    $endgroup$
    – user593746
    Dec 6 '18 at 11:19






  • 2




    $begingroup$
    Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
    $endgroup$
    – Eric
    Dec 6 '18 at 15:40








  • 1




    $begingroup$
    The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
    $endgroup$
    – ancientmathematician
    Dec 6 '18 at 16:34










  • $begingroup$
    My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
    $endgroup$
    – user371231
    Dec 6 '18 at 18:46


















  • $begingroup$
    This is true for any non-zero eigenvalue, not just eigenvalue $1$.
    $endgroup$
    – user593746
    Dec 6 '18 at 11:19






  • 2




    $begingroup$
    Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
    $endgroup$
    – Eric
    Dec 6 '18 at 15:40








  • 1




    $begingroup$
    The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
    $endgroup$
    – ancientmathematician
    Dec 6 '18 at 16:34










  • $begingroup$
    My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
    $endgroup$
    – user371231
    Dec 6 '18 at 18:46
















$begingroup$
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
$endgroup$
– user593746
Dec 6 '18 at 11:19




$begingroup$
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
$endgroup$
– user593746
Dec 6 '18 at 11:19




2




2




$begingroup$
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
$endgroup$
– Eric
Dec 6 '18 at 15:40






$begingroup$
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
$endgroup$
– Eric
Dec 6 '18 at 15:40






1




1




$begingroup$
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
$endgroup$
– ancientmathematician
Dec 6 '18 at 16:34




$begingroup$
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
$endgroup$
– ancientmathematician
Dec 6 '18 at 16:34












$begingroup$
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
$endgroup$
– user371231
Dec 6 '18 at 18:46




$begingroup$
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
$endgroup$
– user371231
Dec 6 '18 at 18:46










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
      $endgroup$
      – A.Γ.
      Dec 6 '18 at 12:50










    • $begingroup$
      Both has same rank as $2.$
      $endgroup$
      – neelkanth
      Dec 6 '18 at 13:03










    • $begingroup$
      @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
      $endgroup$
      – Eric
      Dec 6 '18 at 15:47








    • 1




      $begingroup$
      It’s standard notation if we work in matrices
      $endgroup$
      – neelkanth
      Dec 6 '18 at 16:15










    • $begingroup$
      Can you explain why the ranks are same ?
      $endgroup$
      – user371231
      Dec 6 '18 at 16:18











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.






        share|cite|improve this answer









        $endgroup$



        Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 11:17









        A.Γ.A.Γ.

        22.6k32656




        22.6k32656























            1












            $begingroup$

            As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
              $endgroup$
              – A.Γ.
              Dec 6 '18 at 12:50










            • $begingroup$
              Both has same rank as $2.$
              $endgroup$
              – neelkanth
              Dec 6 '18 at 13:03










            • $begingroup$
              @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
              $endgroup$
              – Eric
              Dec 6 '18 at 15:47








            • 1




              $begingroup$
              It’s standard notation if we work in matrices
              $endgroup$
              – neelkanth
              Dec 6 '18 at 16:15










            • $begingroup$
              Can you explain why the ranks are same ?
              $endgroup$
              – user371231
              Dec 6 '18 at 16:18
















            1












            $begingroup$

            As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
              $endgroup$
              – A.Γ.
              Dec 6 '18 at 12:50










            • $begingroup$
              Both has same rank as $2.$
              $endgroup$
              – neelkanth
              Dec 6 '18 at 13:03










            • $begingroup$
              @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
              $endgroup$
              – Eric
              Dec 6 '18 at 15:47








            • 1




              $begingroup$
              It’s standard notation if we work in matrices
              $endgroup$
              – neelkanth
              Dec 6 '18 at 16:15










            • $begingroup$
              Can you explain why the ranks are same ?
              $endgroup$
              – user371231
              Dec 6 '18 at 16:18














            1












            1








            1





            $begingroup$

            As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.






            share|cite|improve this answer











            $endgroup$



            As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 6 '18 at 16:21

























            answered Dec 6 '18 at 12:20









            neelkanthneelkanth

            2,0642928




            2,0642928












            • $begingroup$
              It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
              $endgroup$
              – A.Γ.
              Dec 6 '18 at 12:50










            • $begingroup$
              Both has same rank as $2.$
              $endgroup$
              – neelkanth
              Dec 6 '18 at 13:03










            • $begingroup$
              @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
              $endgroup$
              – Eric
              Dec 6 '18 at 15:47








            • 1




              $begingroup$
              It’s standard notation if we work in matrices
              $endgroup$
              – neelkanth
              Dec 6 '18 at 16:15










            • $begingroup$
              Can you explain why the ranks are same ?
              $endgroup$
              – user371231
              Dec 6 '18 at 16:18


















            • $begingroup$
              It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
              $endgroup$
              – A.Γ.
              Dec 6 '18 at 12:50










            • $begingroup$
              Both has same rank as $2.$
              $endgroup$
              – neelkanth
              Dec 6 '18 at 13:03










            • $begingroup$
              @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
              $endgroup$
              – Eric
              Dec 6 '18 at 15:47








            • 1




              $begingroup$
              It’s standard notation if we work in matrices
              $endgroup$
              – neelkanth
              Dec 6 '18 at 16:15










            • $begingroup$
              Can you explain why the ranks are same ?
              $endgroup$
              – user371231
              Dec 6 '18 at 16:18
















            $begingroup$
            It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 12:50




            $begingroup$
            It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
            $endgroup$
            – A.Γ.
            Dec 6 '18 at 12:50












            $begingroup$
            Both has same rank as $2.$
            $endgroup$
            – neelkanth
            Dec 6 '18 at 13:03




            $begingroup$
            Both has same rank as $2.$
            $endgroup$
            – neelkanth
            Dec 6 '18 at 13:03












            $begingroup$
            @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
            $endgroup$
            – Eric
            Dec 6 '18 at 15:47






            $begingroup$
            @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
            $endgroup$
            – Eric
            Dec 6 '18 at 15:47






            1




            1




            $begingroup$
            It’s standard notation if we work in matrices
            $endgroup$
            – neelkanth
            Dec 6 '18 at 16:15




            $begingroup$
            It’s standard notation if we work in matrices
            $endgroup$
            – neelkanth
            Dec 6 '18 at 16:15












            $begingroup$
            Can you explain why the ranks are same ?
            $endgroup$
            – user371231
            Dec 6 '18 at 16:18




            $begingroup$
            Can you explain why the ranks are same ?
            $endgroup$
            – user371231
            Dec 6 '18 at 16:18


















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            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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