Prove that if $A$ is nilpotent, then the "jumps" in the nullity of $A^i$ are decreasing. 
Let $A$ be a nilpotent matrix of order $n$ and define $$\forall 1\leq i\leq n-1\hspace{0.6cm} a_i= \text{null} A^{i+1} - \text{null} A^{i}$$
Show that $(a_i)_{i=1}^{n-1}$ is a decreasing sequence. (By $\text{null} A$, we mean the dimension of the kernel of $A$.)

Now let me explain the problem a little more casually. Assume we are working on the vector space $\mathbb{F}^m$, where $\mathbb{F}$ is a field. In the sequence
$$\{0\}\subset \text{ker} A\subset \text{ker} A^2\subset \cdots \subset \text{ker} A^n=\mathbb{F}^m$$ the dimension of each of these subspaces is increasing from $0$ to $m$. So, let's say from $\{0\}$ to $\text{ker} A$ the dimension goes up by $a_1$, from $\text{ker} A$ to $\text{ker} A^2$ the dimension goes up by $a_2$, meaning that $\text{dim ker} A^2= a_1+a_2$, and so on. What I'm trying to prove is that these "jumps" are decreasing.
I've been stuck on this for hours now; I'm not even sure if it's correct or not. I'm not sure how to approach it.
 A: If $A^n = 0$ but $A^{n-1} \neq 0$, then the sequence $$0 \subsetneq \ker A\subsetneq \ker A^2\subsetneq \cdots \subsetneq \ker A^{n-1} \subsetneq V$$is strictily increasing, and $A$ restricted to $\ker A^{i+1}$ takes values in $\ker A^i$. Therefore, $A$ itself induces linear maps $$\frac{\ker A^{i+1}}{\ker A^i} \to \frac{\ker A^i}{\ker A^{i-1}}.$$The thing is that each such map is injective: if $x+\ker A^i$ (with $x\in \ker A^{i+1}$) gets sent to $Ax+\ker A^{i-1} = 0 +\ker A^{i-1}$, then $Ax\in \ker A^{i-1}$, so that $x\in \ker A^i$ and $x+\ker A^i = 0+\ker A^i$. This means that $$a_i = \dim \frac{\ker A^{i+1}}{\ker A^i} \leq \dim\frac{\ker A^i}{\ker A^{i-1}} = a_{i-1},$$as required.
A: Since $A$ is nilpotent, its only eigenvalue is $0$, the characteristic polynomial splits, and so $A$ has a Jordan canonical form $J$.
If $\beta$ is a Jordan canonical basis for $J$, consisting of cycles $\gamma_1,\ldots,\gamma_q$ of generalized eigenvectors of lengths $p_1\geq p_2\geq\cdots\geq p_q$, with
$$\gamma_i = [A^{p_i-1}v_i, \ldots,Av_i,v_i]$$
then a basis vector in $\beta$ lies in $\ker(A^j)$ if and only if it is among the first $j$ vectors in one of the cycles. Therefore, we have:

*

*$\mathrm{nullity}(A)$ is the number of cycles.

*$\mathrm{nullity} (A^2)-\mathrm{nullity}(A)$ is the number of cycles of length at least $2$.

*$\mathrm{nullity}(A^3)-\mathrm{nullity}(A^2)$ is the number of cycles of length at least $3$.

And so on. In general, $\mathrm{nullity}(A^{k}) - \mathrm{nullity}(A^{k-1})$ is the number of cycles of length at least $k$.
This sequence is of course non-increasing, since you cannot have more cycles of length at least $k+1$ than cycles of length at least $k$.
The sequence need not be strictly decreasing, however.  For example, the matrix
$$A=\left(\begin{array}{cccc}
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}\right)$$
has $A^2=0$, so $a_1=2$, $a_2=2$.
