How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker{T^{k-1}}$ and the corresponding canonical forms?

Question

Let $V$ be $5$-dimension vector space, and $T:\ V\rightarrow V$ a nilpotent linear transformation of order (index) $k$ where $1\le k\le 5$. How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker{T^{k-1}}$ and the corresponding canonical forms?

In my personal opinion, I feel like all dims are possible. For example $T=I$, then $\ker{T}=\ker{T^2}=...=\ker{T^{k-1}}=\{0\}$. So dimension is $1$. But I have no idea for the canonical form.

As for other dimensions, since it's in increasing order, if $\dim(\ker{T^k})=n$, then $\dim(\ker{T^{k-1}})\le n$.

The problem is I have no firm proof of this and I don't know how to provide general canonical form. Could someone give any insight?

Answer 1

You really should specify what you mean by "canonical form". I will assume that you meant rational canonical form since you did not specify that you were dealing with a complex vector space. In this particular problem, because of the nilpotency, these are also the Jordan forms, but that is just luck, the distinction of which canonical form you are talking about is important generally.

Also, I will leave the computation of the dimensions of the kernels to you. With the canonical forms written down, it will be straightforward.

$k=1:$ $T$ must be the zero transformation. The zero matrix is its canonical form.

$k=2:$ The minimal polynomial is $x^2$. The invariant factors are then $x,x,x,x^2$ or $x,x^2,x^2$. The corresponding rational canonical forms are respectively, $$\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \textrm{ or } \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}_.$$

$k=3:$ The minimal polynomial is $x^3$. The invariant factors are then $x,x,x^3$ or $x^2,x^3$. The corresponding rational canonical forms are respectively, $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \textrm{ or } \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}_.$$

$k=4:$ The minimal polynomial is $x^4$. The invariant factors are $x,x^4$. The corresponding rational canonical form is, $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}_.$$

$k=5:$ The minimal polynomial is $x^5$. The invariant factor is $x^5$. The corresponding rational canonical form is, $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}_.$$

Answer 2

Notation : We note $K_i=\ker (T^i)$.

As you said we have $\{0\}=K_0 \subset K_1 \subset K_2 \subset \dots \subset K_n=\mathbb{C}^n$. But we have a stronger result.

Proposition : $\forall 1 \le i \le n-1, \dim(K_i) - \dim(K_{i-1}) \geq \dim(K_{i+1}) - \dim(K_i)$.

Proof : Consider the morphisms :

$\begin{array}{ccccc} & K_{i+1} & \xrightarrow{\nu} & K_i & \xrightarrow{\pi_i} & K_i /K_{i-1}\\ & X & \mapsto & TX & \mapsto & \overline{TX}\\ \end{array}$

$\ker(\pi_i \circ \nu)=\nu^{-1}(\pi_i^{-1}(\{0\}))=\nu^{-1}(K_{i-1})=K_i$, so there is an injection from $K_{i+1} /K_{i}$ to $K_i /K_{i-1}$. So $\dim(K_{i+1} /K_{i}) \le \dim(K_i /K_{i-1})$. And the propositions follows.

To a nilpotent matrice you can associate a Young tableau such as the bottom line has $\dim K_1 - \dim K_0$ elemnts, the second line has $\dim K_2 - \dim K_1$ elements and so on. You can prove that two nilpotent matrices are similar iff thay have the same Young tableau. And they are similar to a normal form : a matrix with Jordan blocs, and the dimensions of the Jordan blocs are given by the dimensions of the rows of the Young tableau.

Here is an example : If $A\in M_{10}(\mathbb{C})$, is nilpotent of order $3$ such as $\dim(K_1)=2$ and $\dim(K_2)=5$. $\dim(K_3)-\dim(K_2)=5,\dim(K_2)-\dim(K_1)=3,\dim(K_1)-\dim(K_0)=2$. The Young tableau has the shape:

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So the rows are of sizes : $3,3,2,1,1$. So $A$ is similar to $$\begin{bmatrix} J_3 & & & & \\ & J_3 & & & \\ & & J_2 & & \\ & & & J_1 & \\ & & & & J_1 \end{bmatrix}$$ Where $J_k$ is a Jordan block of size $k$.