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Let V is a finite dimensional vector space over C and T be a linear operator on V . How to prove T has an invariant subspace of dimension k for each k = 1,2, . . . ,dimV .

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It can be solved by Jordan form. I just give a hint.

Let $\{e_i\}$ is a basis of $V$. Put $J=\lambda I+N$, where $I$ is the identity map, and $N$ is the nilpotent linear map such that sent $e_i$ to $e_{i+1}$(if $i+1>\dim V$, then 0). Every subspace of the form $<a_{k+1}e_{k+1}+,...,+a_ne_n>$ is an invariant subspace of $\dim n-k$.

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  • $\begingroup$ How to take $N$ $e_i$ to $e_i+1$ $\endgroup$ – Shona May 7 '16 at 20:37
  • $\begingroup$ Can you pls elaborate ?? $\endgroup$ – Shona May 8 '16 at 6:30
  • $\begingroup$ Maybe you can try to write the matrix of the linear map $\lambda I+N$ under the basis $<e_i>$. It will show that the matrix is standard Jordon matrix. $\endgroup$ – Xerxes May 8 '16 at 8:35
  • $\begingroup$ Won't it give only $\lambda$ 's since $N$ is nilpotent then its dimension would be the dimension of $I$ $\endgroup$ – Shona May 8 '16 at 8:45
  • $\begingroup$ How to do it ?? $\endgroup$ – Shona May 9 '16 at 9:12

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