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Let $N_{i}=\{A \in M_{n\times n}(k) \mid A^{i}=0\}$ that is, $N_i$ is the set of nilpotent matrix with order $i$. Then we get the chain $N_1 \subset N_2 \subset \cdots \subset N_n$. Is $N_i \subsetneqq N_{i+1}$ for all $i$ true?

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Six of your questions got answers, but you accepted none of them. Why? – Did Nov 1 '11 at 10:26
up vote 1 down vote accepted

The following exercises will enhance your understanding on the question that you have asked:

$\textbf{(1)}$ Prove that if there exists a positive integer $k$ such that null $A^k$ = null $A^{k+1}$, then null $A^n$ = null $A^{n+1}$ for all integers $n \geq k$.

$\textbf{(2)}$ Prove that if $n =$ dim $V$, and $A$ an operator on $V$ then null $A^n = $ null $A^{n+1} =$ null $A^{n+2} = ...$

Hint: Use the fact that $V$ is of finite dimension and that no subspace of $V$ can be of a larger dimension than $V$.

$\textbf{(3)}$ If $A$ is a nilpotent operator on a complex vector space $V$, prove that $A^{\text{dim} V}$ = 0.

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Yes for $i<n$. Jordan normal form makes it easy to find an example of a matrix with $A^i\neq 0$, but $A^{i+1}=0$.

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