# 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.

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.

• You have my gratitude. This was what I was looking for. Is there any book that has results like these which could come in handy while solving problems in Linear Algebra?? Commented Sep 23, 2015 at 8:28
• To be honest, I cannot remember to have seen that result in a text book. It is just something, that comes into your mind, if you understand the mechanism behind the Jordan form: Take a Jordan form matrix and consider its powers. Anytime you increase the power, each block with at least one $1$ remaining, contributes to the increase of dimension of the kernel by exactly one. A block of starting size $m$ will be eventually vanishing at the $m$-th power. So the amount of "blocks alive", which contribute the dimensional increase of the kernel, only gets lower and lower...
– MooS
Commented Sep 23, 2015 at 8:52
• That was very neatly put. I guess I have to rely on my understanding rather than reach for a book for any and every result. Commented Sep 23, 2015 at 9:30
• By the way: It might be worth while coming up with a rigorous proof of the constructive result, which is of course a little bit technical. You should make sure to point out, where we really need the difference sequence to be decreasing in order to make the proof work.
– MooS
Commented Sep 23, 2015 at 10:01

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.

• Thanks for your trouble. But I have almost no knowledge of modules as such. Could you explain using a vector space over the reals approach??I guess that would follow as a special case of what you said. But it would still be very helpful. Thanks again Commented Sep 23, 2015 at 6:27
• Hm, okay. I imagine that my argument can be rephrased in terms of generalized eigenvectors. Have you dealt with these before? If so, I'll think about it some more tomorrow. Commented Sep 23, 2015 at 6:37
• Another more basic approach would be to write down all the possible Jordan canonical forms $J$ for $T$ and then compute the kernel of $J, J^2, \ldots$ for each one. It will require a bit more computation, but there aren't that many possibilities for $J$. Commented Sep 23, 2015 at 6:40
• Yes I do know about generalised eigenvectors. Will give it a go myself again before you post your answer. I was looking for some method that doesnt involve too much computation as this question propped up in a timed exam. But anything helps. Thank you. Commented Sep 23, 2015 at 8:04
• I apologise for not seeing your edit at all. I dont why only your comments were visible. The nullity would be 2 and that would give me my answer. Commented Sep 23, 2015 at 8:50