Proof about an endomorphism of a finite-dimensional vector space











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Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,



$$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$



I don't know exactly how to start this proof. Could you help me? Thanks in advance!






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    up vote
    2
    down vote

    favorite












    Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,



    $$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$



    I don't know exactly how to start this proof. Could you help me? Thanks in advance!






    linear-algebra







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,



      $$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$



      I don't know exactly how to start this proof. Could you help me? Thanks in advance!






      linear-algebra







      share|cite|improve this question















      Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,



      $$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$



      I don't know exactly how to start this proof. Could you help me? Thanks in advance!






      linear-algebra




      linear-algebra






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      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 21 at 18:07









      Monstrous Moonshiner

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      asked Nov 21 at 17:40









      Gibbs

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          First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.



          The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.






          share|cite|improve this answer























          • Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
            – Gibbs
            Nov 21 at 18:43






          • 1




            @Gibbs That is correct
            – Monstrous Moonshiner
            Nov 21 at 18:54











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          1 Answer
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          oldest

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          active

          oldest

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          up vote
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          down vote



          accepted










          First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.



          The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.






          share|cite|improve this answer























          • Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
            – Gibbs
            Nov 21 at 18:43






          • 1




            @Gibbs That is correct
            – Monstrous Moonshiner
            Nov 21 at 18:54















          up vote
          1
          down vote



          accepted










          First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.



          The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.






          share|cite|improve this answer























          • Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
            – Gibbs
            Nov 21 at 18:43






          • 1




            @Gibbs That is correct
            – Monstrous Moonshiner
            Nov 21 at 18:54













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.



          The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.






          share|cite|improve this answer














          First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.



          The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 18:54

























          answered Nov 21 at 18:03









          Monstrous Moonshiner

          2,25511337




          2,25511337












          • Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
            – Gibbs
            Nov 21 at 18:43






          • 1




            @Gibbs That is correct
            – Monstrous Moonshiner
            Nov 21 at 18:54


















          • Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
            – Gibbs
            Nov 21 at 18:43






          • 1




            @Gibbs That is correct
            – Monstrous Moonshiner
            Nov 21 at 18:54
















          Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
          – Gibbs
          Nov 21 at 18:43




          Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
          – Gibbs
          Nov 21 at 18:43




          1




          1




          @Gibbs That is correct
          – Monstrous Moonshiner
          Nov 21 at 18:54




          @Gibbs That is correct
          – Monstrous Moonshiner
          Nov 21 at 18:54


















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