possible dimensions of a nilpotent endomorphism of a K-Vectorspace

Question

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?

Answer 1

If you know the theory of the Jordan Form this is reasonably straightforward. (But I suspect that you are expected to do the problem bare-handed as it were.)

As $N$ is nilpotent we have that $N^k=0$ for some $k$, and as $\dim V=5$ we see (by Cayley-Hamilton) that $k\leqslant 5$, and $m_A(X)=X, X^2, X^3, X^4, \text{ or } X^5$.

The eigenvalues of $N$ are all $0$, and by the general theory $N$ is similar to one of

$$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}. $$ The ranks of these are visibly $0,1,2,2,3,3,4$.

As for $N^2$ we have the following (in the same order) $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix} $$ whose ranks are $0,0,0,1,1,2,3$.

Now we get for $N^3$ $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix} $$ whose ranks are $0,0,0,0,0,1,2$.

For $N^4$ we have $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{pmatrix} $$ whose ranks are $0,0,0,0,0,0,1$.

For all $k\geqslant 5$ we have $N^5=0$ whose rank is $0$.

By the Rank-Nullity Theorem you can now write down the possible sequences $(\dim\ker N^k)_{k=0}^{\infty}$

Answer 2

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$~n$.