Nilpotent Operators Linear Algebra












1














Let $V$ be an $n$-dimensional vector space over an arbitrary field $K$, and let
$T_1, dots , T_n : V rightarrow V$ be pairwise commuting nilpotent operators on V



(a) Show the composition $T_1 T_2 · · · T_n = 0$.



(b) Does this still hold if we drop the hypothesis that the $T_i$ are pairwise commuting?



What I have tried:
I know that for any $T_i$, $ker({T_i}^n) = ker({T_i}^{n+1}) = cdots$. This implies that ${T_i}^n = 0$ for any $i$. I also know that raising each $T_i$ to a power of itself lessens the dimension of the kernel of such a composition. Any help?










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    1














    Let $V$ be an $n$-dimensional vector space over an arbitrary field $K$, and let
    $T_1, dots , T_n : V rightarrow V$ be pairwise commuting nilpotent operators on V



    (a) Show the composition $T_1 T_2 · · · T_n = 0$.



    (b) Does this still hold if we drop the hypothesis that the $T_i$ are pairwise commuting?



    What I have tried:
    I know that for any $T_i$, $ker({T_i}^n) = ker({T_i}^{n+1}) = cdots$. This implies that ${T_i}^n = 0$ for any $i$. I also know that raising each $T_i$ to a power of itself lessens the dimension of the kernel of such a composition. Any help?










    share|cite|improve this question



























      1












      1








      1







      Let $V$ be an $n$-dimensional vector space over an arbitrary field $K$, and let
      $T_1, dots , T_n : V rightarrow V$ be pairwise commuting nilpotent operators on V



      (a) Show the composition $T_1 T_2 · · · T_n = 0$.



      (b) Does this still hold if we drop the hypothesis that the $T_i$ are pairwise commuting?



      What I have tried:
      I know that for any $T_i$, $ker({T_i}^n) = ker({T_i}^{n+1}) = cdots$. This implies that ${T_i}^n = 0$ for any $i$. I also know that raising each $T_i$ to a power of itself lessens the dimension of the kernel of such a composition. Any help?










      share|cite|improve this question















      Let $V$ be an $n$-dimensional vector space over an arbitrary field $K$, and let
      $T_1, dots , T_n : V rightarrow V$ be pairwise commuting nilpotent operators on V



      (a) Show the composition $T_1 T_2 · · · T_n = 0$.



      (b) Does this still hold if we drop the hypothesis that the $T_i$ are pairwise commuting?



      What I have tried:
      I know that for any $T_i$, $ker({T_i}^n) = ker({T_i}^{n+1}) = cdots$. This implies that ${T_i}^n = 0$ for any $i$. I also know that raising each $T_i$ to a power of itself lessens the dimension of the kernel of such a composition. Any help?







      nilpotence






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      edited Dec 10 '18 at 20:50

























      asked Dec 9 '18 at 23:21







      user624358


































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