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 $n>r_i$?