📂Linear Algebra

Null Space of Power Maps

Theorem¹

$n$ Let’s say that the linear transformation $T : V \to V$ on the dimension vector space is nilpotent.

$$T^{p} = T_{0}$$

Here, $T_{0}$ is the zero transformation. Let’s call $N(T)$ the null space of $T$. Then, the following holds:

1. For all $i \in \mathbb{N}$, it is $N(T^{i}) \subset N(T^{i+1})$.

2. For $1 \le i \le p-1$, there exists a sequence basis $N(T^{i})$ of $\beta_{i}$ such that the following holds. $$\beta_{i} \subset \beta_{i+1}$$

3. Let’s call the sequence basis obtained by the method of 2. as $N(T^{p}) = V$. Then, the matrix representation $\begin{bmatrix} T \end{bmatrix}_{\beta}$ is an upper triangular matrix.

4. The characteristic polynomial of $T$ is $(-1)^{n}t^{n}$. Therefore, $T$ can be decomposed, and the eigenvalues of $T$ are only $0$.

Explanation

1. Since the null space is a subspace of $V$, $N(T^{i})$s are increasingly larger subspaces of $V$. There are up to $p-1$ proper subspaces. $$\left\{ \mathbf{0} \right\} \le N(T^{1}) \le N(T^{2}) \le \cdots \le N(T^{p}) = V$$

2. The fact that the eigenvalues of nilpotent are only $0$ can easily be shown with another proof.

Proof

1.

If $T^{i}\mathbf{v} = \mathbf{0}$, then it is $T^{i+1}\mathbf{v} = TT^{i}\mathbf{v} = T\mathbf{0} = \mathbf{0}$.

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2.

Choose one of the sequence bases of $N(T^{1})$ as $\beta_{1}$ randomly. Then, based on the result of 1., you can get $\beta_{2}$ that becomes a sequence basis of $N(T^{2})$ by expanding $\beta_{1}$. Repeating this to get $\beta_{i}$ completes the proof.

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