# Question about kernel of the restriction of a linear operator

I'm studying linear algebra and I am trying to answer a question I asked myself.

Suppose $T:V\rightarrow V$ is a linear operator on a finite dimensional vector space $V$ over an algebraically closed field $K$.

Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of $T$. Let $$m_T(x)=(x-\lambda_1)^{r_1}\dots(x-\lambda_n)^{r_n}$$ be the minimal polynomial of $T$.

By the Primary Decomposition Theorem:

$$V=\ker(T-\lambda_1 I_V)^{r_1}\oplus\dots\oplus\ker(T-\lambda_nI_V)^{r_n}$$

Let $V_i=\ker(T-\lambda_i I_V)^{r_i}$. Then $T$ is $V_i$ invariant. Let $T|V_i=T_i$.Then $T_i$ is nilpotent and it's minimal polynomial is $(x-\lambda_i)^{r_i}$.

It's easy to show that for every $n\leq r_i$, $\ker (T_i-\lambda_i I_{V_i})^n=\ker (T-\lambda_iI_V)^n$. My question is: Is it also true for $n>r_i$?

## 1 Answer

The proof of the primary decomposition theorem or a direct argument shows that the sum

$$\ker(T - \lambda_1 I_V)^{k_1} + \dots + \ker(T - \lambda_n I_V)^{k_n}$$

is always a direct sum $$\ker(T - \lambda_1 I_V)^{k_1} \oplus \dots \oplus \ker(T - \lambda_n I_V)^{k_n}$$ for any $k_1, \dots, k_n \geq 0$ and distinct $\lambda_1, \dots, \lambda_n$ (only the sum doesn't necessarily add up to the whole of $V$). In other words, generalized eigenvectors that are associated with distinct eigenvalues are linearly independent.

Now note that $T_i$ is not nilpotent but $T_i - \lambda_i I_{V_i}$ is with $(T_i - \lambda_i I_{V_i})^{r_i} = 0_{V_i}$ so if $n \geq r_i$ then $\ker(T_i - \lambda_i I_{V_i})^{n} = V_i = \ker(T - \lambda_i I_V)^{r_i}$. Thus, it is sufficient to show that $V_i = \ker(T - \lambda_i I_V)^{r_i} = \ker(T - \lambda_i I_V)^n$ for $n \geq r_i$. To see this, let $w \in \ker(T - \lambda_i I_V)^n$ and decompose it as $v = v_1 + \dots + v_n$ using the direct sum decomposition guaranteed by the primary decomposition theorem. But then

$$v - v_i + \sum_{j \neq i} v_j = 0$$

where $v - v_i$ and each $v_j$ are generalized eigenvectors that are associated with distinct eigenvalues and so $v = v_i$ and hence $v \in V_i$.