How to prove that $ker(T^k+1)=ker(T^k+2)$ when $ker(T^k)=ker(T^k+1)$. [closed]

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If $ker(T^k)=ker(T^k+1)$, then show that $ker(T^k+1)=ker(T^k+2)$.



If $operatornameim(T^k)=operatornameim(T^k+1)$, then show that $operatornameim(T^k+1)=operatornameim(T^k+2)$.




vector-spaces




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closed as off-topic by Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey Aug 21 at 14:26


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey
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  • Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
    – Jneven
    Aug 21 at 11:21










  • What have you done so far? I am also sure that this question was already answered on MSE.
    – amsmath
    Aug 21 at 11:21










  • When is this due?
    – Did
    Aug 21 at 11:47














up vote
-2
down vote

favorite












If $ker(T^k)=ker(T^k+1)$, then show that $ker(T^k+1)=ker(T^k+2)$.



If $operatornameim(T^k)=operatornameim(T^k+1)$, then show that $operatornameim(T^k+1)=operatornameim(T^k+2)$.




vector-spaces




share|cite|improve this question














closed as off-topic by Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey Aug 21 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
    – Jneven
    Aug 21 at 11:21










  • What have you done so far? I am also sure that this question was already answered on MSE.
    – amsmath
    Aug 21 at 11:21










  • When is this due?
    – Did
    Aug 21 at 11:47












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











If $ker(T^k)=ker(T^k+1)$, then show that $ker(T^k+1)=ker(T^k+2)$.



If $operatornameim(T^k)=operatornameim(T^k+1)$, then show that $operatornameim(T^k+1)=operatornameim(T^k+2)$.




vector-spaces




share|cite|improve this question














If $ker(T^k)=ker(T^k+1)$, then show that $ker(T^k+1)=ker(T^k+2)$.



If $operatornameim(T^k)=operatornameim(T^k+1)$, then show that $operatornameim(T^k+1)=operatornameim(T^k+2)$.





vector-spaces





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 21 at 11:30









Bernard

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111k635103










asked Aug 21 at 11:15









mathnewbie

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closed as off-topic by Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey Aug 21 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey Aug 21 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, Jyrki Lahtonen, amWhy, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
    – Jneven
    Aug 21 at 11:21










  • What have you done so far? I am also sure that this question was already answered on MSE.
    – amsmath
    Aug 21 at 11:21










  • When is this due?
    – Did
    Aug 21 at 11:47
















  • Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
    – Jneven
    Aug 21 at 11:21










  • What have you done so far? I am also sure that this question was already answered on MSE.
    – amsmath
    Aug 21 at 11:21










  • When is this due?
    – Did
    Aug 21 at 11:47















Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
– Jneven
Aug 21 at 11:21




Possible duplicate of given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.
– Jneven
Aug 21 at 11:21












What have you done so far? I am also sure that this question was already answered on MSE.
– amsmath
Aug 21 at 11:21




What have you done so far? I am also sure that this question was already answered on MSE.
– amsmath
Aug 21 at 11:21












When is this due?
– Did
Aug 21 at 11:47




When is this due?
– Did
Aug 21 at 11:47










1 Answer
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It's easy to see that $ker(T^k+1)subset ker(T^k+2)$ so just you need the other inclusion. Let $xin ker(T^k+2)$ i.e. $T^k+2(x)=0$, hence $T(x)in ker(T^k+1)=ker(T^k)$ which gives the desired result $T^k+1(x)=T^k(T(x))=0$, i.e. $xin ker(T^k+1)$.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote













    It's easy to see that $ker(T^k+1)subset ker(T^k+2)$ so just you need the other inclusion. Let $xin ker(T^k+2)$ i.e. $T^k+2(x)=0$, hence $T(x)in ker(T^k+1)=ker(T^k)$ which gives the desired result $T^k+1(x)=T^k(T(x))=0$, i.e. $xin ker(T^k+1)$.






    share|cite|improve this answer
























      up vote
      1
      down vote













      It's easy to see that $ker(T^k+1)subset ker(T^k+2)$ so just you need the other inclusion. Let $xin ker(T^k+2)$ i.e. $T^k+2(x)=0$, hence $T(x)in ker(T^k+1)=ker(T^k)$ which gives the desired result $T^k+1(x)=T^k(T(x))=0$, i.e. $xin ker(T^k+1)$.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        It's easy to see that $ker(T^k+1)subset ker(T^k+2)$ so just you need the other inclusion. Let $xin ker(T^k+2)$ i.e. $T^k+2(x)=0$, hence $T(x)in ker(T^k+1)=ker(T^k)$ which gives the desired result $T^k+1(x)=T^k(T(x))=0$, i.e. $xin ker(T^k+1)$.






        share|cite|improve this answer












        It's easy to see that $ker(T^k+1)subset ker(T^k+2)$ so just you need the other inclusion. Let $xin ker(T^k+2)$ i.e. $T^k+2(x)=0$, hence $T(x)in ker(T^k+1)=ker(T^k)$ which gives the desired result $T^k+1(x)=T^k(T(x))=0$, i.e. $xin ker(T^k+1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 21 at 11:21









        user296113

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