Prove that dim$N(A^i)=i$ if $A$ is nilpotent Let $A\in\mathbb{C^{n\times n}}$, $A$ is nilpotent such that dim$N(A)=1$, where $N(A)$ is a null space.
$\qquad\qquad\qquad\qquad\qquad$ Prove that dim$N(A^i)=i$ for all $i=1,2,\ldots,n.$ 
Looks simple but not obvious. We know that $1=$ dim$N(A)\leq$ dim$N(A^2)\leq\ldots\leq$ dim$N(A^n)=n$. We also can find that dim$N(A^{n-1})\neq n$.
 A: HINT:
Let $m$ minimal so that $A^m=0$, then take $v$ so that $A^{m-1}v \ne 0$. Consider the vectors $v_m = v$, $v_k =A^{m-k}v$, $\ldots$, $v_1=A^{m-1}v$. Show by induction on $k$ that $N(A^k)$ is generated by $v_1$ , $\ldots$, $v_k$. 
A: Instead of working in $\mathbb{C}^n$, I'm going to work in an arbitrary vector space $V$ of dimension $n$.
Let $\dim N(A^i) = n_i$. As you say, it is clear that $n_{i - i} \leq n_i$ for all $i$. First we will show that $n_{i -1 } < n_i$ for all $i$ such that $n_i < n $. Suppose not. Then $n_{k-1} = n_k$ for some $k$ such that $n_k < n$. That is, $$\dim N(A^{k-1}) = \dim N(A^k) = n_k < n$$ 
Since $ N(A^{i-1}) \subseteq N(A^i)$ for all $i$ (in particular for $k$), this means that $N(A^{k -1}) = N(A^k)$. Take $m > k$. If $v \in N(A^m)$, then $A^{m}v = 0$, so $A^k A^{m - k}v = 0$. This means $A^{m - k}v \in N(A^k)$, which implies $A^{m  - k} v \in N(A^{k -1})$. Thus $A^{k - 1} A^{m - k} v = 0$, so $A^{m - 1} v = 0$, i.e. $v \in N(A^{m - 1})$. Thus, $N(A^{m -1}) = N(A^m)$. Therefore, by induction, $N(A^{k - 1}) = N(A^k) = \cdots = N(A^m)$ for all $m > k$. This means $n_m = n_k < n$ for all $m \geq k$. But, by rank-nullity, this means $Im(A^k) \neq \{0\}$ for all $k$. In other words, $A^k \neq 0$ for all $k$, a contradiction. 
Now we show that $n_{j + 1} - n_j \leq n_{j} - n_{j - 1}$ for all $j \geq 1$. Consider 
$$
\phi : Im(A^{j-1}) / Im(A^{j}) \to Im(A^{j}) / Im(A^{j+1})
$$
given by $\phi(v + Im(A^{j})) = Av + Im(A^{j+ 1})$. This map is clearly linear. To see that this map is well-defined, take $w \in Im(A^{j})$. Then $Aw \in Im(A^{j+1})$, so
\begin{align}
\phi(v + w + Im(A^{j})) &= A(v + w) + Im(A^{j+1}) 
= Av + Aw + Im(A^{j+1}) 
\\&= Av + Im(A^{j+1}) = \phi(v + Im(A^j))
\end{align}
Moreover, this map is surjective. If $w + Im(A^{j+1}) \in Im(A^{j}) / Im(A^{j+1})$, then $w = A^{j} x$ for some $x \in V$. Thus
$$
\phi(A^{j-1} x + Im(A^{j})) = A^{j } x + Im(A^{j+1}) = w + Im(A^{j+1})
$$
Since $\phi$ is surjective, we may conclude
\begin{align}
\dim Im(A^{j-1}) &- \dim Im(A^{j}) = \dim Im(A^{j-1}) / Im(A^{j})  
\\&\geq \dim Im(A^{j}) / Im(A^{j+1}) 
= \dim Im(A^{j}) - \dim Im(A^{j+1})
\end{align}
By rank-nullity $\dim Im(A^j) = n - n_j$, so
$$
(n - n_{j-1}) - (n - n_{j}) \geq (n - n_{j}) - (n - n_{j+1})
$$
On simplifying, we find $n_j - n_{j-1} \geq n_{j + 1} - n_j$, which is what we wanted to show.
Finally, $n_0 = 0$ and $n_1 = \dim N(A) = 1$, so $1 \geq n_{j+1} - n_{j}$ for all $j \geq 0$. On the other hand, $n_{j + 1} - n_j > 0$ for all $n_{j + 1} < n$, so $n_{j + 1} - n_j = 1$ for all $n_{j + 1} < n$. That is, $n_{j + 1} = 1 + n_j$ for all $n_{j + 1} < n$ and $n_1 = 1$, from which we may conclude $n_j = j$ for all $j < n$. Hopefully it is clear that $n_k = n$ for all $k \geq n$. 
This answer is long and strictly less elegant than the accepted answer, but I began writing it before that answer was posted. I'm going to leave it up since it proves two facts I couldn't easily find compelling proofs of elsewhere online. The first is that for any nilpotent operator $A$, $\dim A^i$ strictly increases until it reaches $n$, at which point it remains constant.  The second is that, for any operator $A$, if $\dim A^j = n_j$, then $n_{j + 1} - n_j \leq n_{j} - n_{j -1}$ for all $j \geq 1$. 
