possible dimensions of a nilpotent endomorphism of a K-Vectorspace I have the following exercise in linear algebra that I think I know what they're looking for, however have not really got an idea which theorems to use / how to approach this.
Let $V$ be a 5-dimensional $K$-Vectorspace, and $N$ a nilpotent endomorphism of $V$. For $r \in \mathbb{N}$, we set $V^{(r)} =$ Ker$(N^{\circ r})$ and $d_r =$ dim$V^{(r)}$. What are the possible sequences of integers that one may get in this way$?$ For each possible sequence, give an example of a nilpotent endomorphism $N$ realising it.
From my script I know that Matrix $A$ is nilpotent if $A^k = 0$ for some integer $k$.
So if I understood this correctly, we are looking for possible sequences for dim(Ker$(N^{\circ r})$).
My guess is that based off of the definition of a nilpotent matrix, the sequence will always end with $0$. However I am not sure if there is a theorem to calculate the first elements of the sequence.
How can I approach calculating these possible sequences?
 A: In general, there are a few important this to know about the sequence $(\dim\ker(N^k))_{k\in\Bbb N}$

*

*It starts at $0$ at $k=0$ (since $N^0=I$ by definition of exponentiation);

*It ultimately stabilises at the value $n=\dim(V)$, the size of $N$ (since $N^k=0$ ultimately);

*It is weakly increasing (since $N^kv=0$ implies $N^{k+1}v=0$, so $\ker(N^k)\subseteq\ker(N^{k+1})$);

*In is concave, meaning the sequence of non-negative differences $(\dim\ker(N^{k+1})-\dim\ker(N^k))_{k\in\Bbb N}$ is weakly decreasing; in particular when the difference becomes$~0$ (the original sequence has two equal consecutive terms) it remains$~0$ (the original sequence becomes stationary).

The final point is slightly more difficult to show: first, applying $N$ to vectors maps the subspace $\ker(N^{k+1})$ to $\ker(N^k)$ for every $k\in\Bbb N$, second, this induces a map of quotient spaces $\ker(N^{k+2})/\ker(N^{k+1})\to\ker(N^{k+1})/\ker(N^k)$ that is well defined, and third, that induced (linear) map is injective, which gives the desired inequality.
Thus the sequence of differences is a weakly decreasing sequence of non-negative integers, ultimately becoming $0$, and whose sum is$~n$. Such sequences are called a partition of $n$, and they are very well studied. There are $7$ partitions of $5$, namely (with the trailing zeros omitted) $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1,1,1,1,1)$. The corresponding (original) sequences of dimensions can be found by taking partial sums left to right, starting from the partial sum$~0$ of the empty subsequence, giving respectively

*

*$(0,5,5\ldots)$

*$(0,4,5,5,\ldots)$

*$(0,3,5,5,\ldots)$

*$(0,3,4,5,5,\ldots)$

*$(0,2,4,5,5,\ldots)$

*$(0,2,3,4,5,5,\ldots)$

*$(0,1,2,3,4,5,5,\ldots)$
It is not hard to see that these sequences can all be realised for appropriate nilpotent matrices, and that this is the case for all$~n$.
