Nilpotent matrix and relation between its powers and dimension of kernels Given a 4x4 matrix $T$ over $\mathbb{R}$ such that $T^4 = 0 $, $k_i = \textsf{dim} Ker(T^i)$, I need to check which of the following sequences, $$k_1\leq k_2 \leq k_3 \leq k_4,$$ is NOT possible : 
$ 1)\; 1\leq 3 \leq 4 \leq 4$
$2) \; 2\leq3\leq4\leq4$
$3) \; 3 \leq 4 \leq 4\leq 4$
$4)\; 2 \leq 4 \leq 4 \leq 4$
The only relevant thing I could recall relating to this is the fact that for nilpotent operators $$ \{0 \} \subset Ker(T) \subset Ker(T^2) \subset \ldots \subset Ker(T^{n-1})$$
But the equality in the choices is putting me off. A hint would be welcome. Thanks in advance.
 A: Consider the possible lists invariant factors
of $T$.  Since $T$ is $4 \times 4$, then its characteristic polynomial is $T^4$.  The invariant factors must divide the characteristic polynomial, hence are all of the form $T^i$.  The characteristic polynomial is the product of all invariant factors, so the list of invariant factors must be $T^{i_1}, T^{i_2}, \ldots, T^{i_t}$, where $i_1, \ldots, i_t$ are positive integers, and $i_1 + \cdots + i_t = 4$.  Moreover, each invariant factor divides the following invariant factor, so we have $i_1 \leq \cdots \leq i_t$.  The possible partitions of $4$ are:
\begin{align*}
4 &= 4\\
4 &= 3+1\\
4 &= 2 + 2\\
4 &= 2 + 1 + 1\\
4 &= 1 + 1 + 1+ 1
\end{align*}
Let $k$ be our base field.  We consider $k^4$ as a $k[x]$-module where $x$ acts as $T$.  By the structure theorem for finitely generated modules over a PID
, then $k^4$ is isomorphic to one of the following $k[x]$-modules. 
\begin{align*}
\frac{k[x]}{(x^4)} &\longleftrightarrow 1, 2, 3, 4\\
\frac{k[x]}{(x^3)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 2, 3, 4, 4\\
\frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x^2)} &\longleftrightarrow 2, 4, 4, 4\\
\frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 3, 4, 4, 4\\
\frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 4, 4, 4, 4
\end{align*}
The numbers on the right are the corresponding dimensions $\dim(\ker(T^j))$ for $j = 1, \ldots, 4$.  These numbers were obtained by considering the action of $x$ on the standard basis.  For instance, $\frac{k[x]}{(x^3)} \cong k \oplus kx \oplus kx^2$ is a $k$-vector space with basis $1,x,x^2$.  Multiplying by $x$ annihilates $x^2$, but not $1$ or $x$ (because we are modding out by $x^3$).  Similarly, multiplying by $x^2$ annihilates $x$ and $x^2$ but not $1$.
EDIT: Alternatively, consider the possible Jordan canonical forms for $T$.  Since $T$ is nilpotent, then its only eigenvalue is $0$, so its Jordan canonical form is
$$
J =
\begin{pmatrix}
0 & * & 0 & 0\\
0 & 0 & * & 0\\
0 & 0 & 0 & *\\
0 & 0 & 0 & 0
\end{pmatrix}
$$
where the $*$ are either $0$ or $1$.  This yields $8$ possibilities for $J$ and with a bit of calculation, one can compute the dimensions of the kernels of $J, J^2, J^3, J^4$ for each $J$.  Actually, it's not so hard to find which list is impossible.  If $\dim(\ker(T)) = 1$, then $T$ has rank $3$, so we must have that its Jordan canonical form is
$$
J =
\begin{pmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{pmatrix} \, .
$$
As you noted in the comments,
$$
J^2 =
\begin{pmatrix}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix} \, .
$$
and which has a $2$-dimensional kernel.  This rules out the first choice.
A: There is a very basic and easy-to-prove result:
Lemma: Let $T$ be an endomorphism on an $n$-dimensional vectorspace. Denote $k_i = \dim \operatorname{Ker} T^i$. Note that $k_0 = 0$, since $T^0$ is defined to be the identity map. The sequence $(k_i)_{i \geq 0}$ is increasing, while the sequence of differences $(k_{i+1}-k_i)_{i \geq 0}$ is decreasing.
(This Lemma immediately rules out your possibility 1) 
The first statement is trivial, since we have $\operatorname{Ker} T^i \subset \operatorname{Ker} T^{i+1}$.
Let us prove the second statement:
Denote $b_i = \dim \operatorname{Im} T^i$. Since we have $b_i+k_i = n$ for all $i$, we have $k_{i+1}-k_i = b_i-b_{i+1}$, so it suffices to show, that the latter sequence decreases. For that, just consider the map
$$T:\operatorname{Im} T^i \to V$$
and use dimension formula: The image is $\operatorname{Im} T^{i+1}$ and the kernel is $\operatorname{Im} T^i \cap \operatorname{Ker} T$, so we get
$$b_i = b_{i+1} + \dim \operatorname{Im} T^i \cap \operatorname{Ker} T$$ or
$$b_i-b_{i+1} = \dim \operatorname{Im} T^i \cap \operatorname{Ker} T$$, which is decreasing, since we have $\operatorname{Im} T^i \supset \operatorname{Im} T^{i+1}$ for all  $i$.

As stated, this rules out something like $0 \leq 1 \leq 3 \leq 4$.
A corollary is the following famous result: Once we have $\dim \operatorname{Ker} T^i = \dim \operatorname{Ker} T^{i+1}$, we have $\dim \operatorname{Ker} T^j = \dim \operatorname{Ker} T^i$ for all $j \geq i$.
We also get the following corollary: Let $T$ be nilpotent with one-dimensional kernel. Then the kernel of $T^j$ is $j$-dimensional for any $j \leq n$.

As a constructive result, let us note that the lemma above is the only restriction for the sequence $(k_i)_{i \geq 0}$:
Let $(k_i)_{i \geq 0}, k_0=0$ be an increasing sequence, such that $(k_{i+1}-k_i)$ is decreasing and eventually stabilizes at zero. Then there exists an $n > 0$ and an endomorphism $T$ on an $n$-dimensional vector space, such that $k_i = \dim \operatorname{Ker} T^i$.
The proof is simply the Jordan form.
In particular your possibilites 2)-4) are all possible.
