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 $\dim nk$.
Xerxes
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How to take $N$ $e_i$ to $e_i+1$ – Shona May 07 '16 at 20:37

Can you pls elaborate ?? – Shona May 08 '16 at 06:30

Maybe you can try to write the matrix of the linear map $\lambda I+N$ under the basis $
$. It will show that the matrix is standard Jordon matrix. – Xerxes May 08 '16 at 08:35 
Won't it give only $\lambda$ 's since $N$ is nilpotent then its dimension would be the dimension of $I$ – Shona May 08 '16 at 08:45

How to do it ?? – Shona May 09 '16 at 09:12