If $K_i=\operatorname{ker}(T^i)$ and $\exists r\in\mathbb{N}:K_r=K_{r+1}$ then prove $K_r=K_{r+i}$ $\forall i\geq 1$ Let V be an n-dimensional vector space over $\mathbb{C}$ and $T:V\rightarrow V$ be a linear transformation. For $i\geq 0$ let $K_i=\operatorname{ker}(T^i)$.
I've shown that $K_i \subseteq K_{i+1}$ and that there exists a non-negative integer $r$ such that $K_r=K_{r+1}$
Can anyone help me prove that $K_r=K_{r+i}$ $\forall i\geq 1$ and therefore $V=K_r \oplus T^r(V)$?
I'm attempting it with induction, clearly it is true for $i=0$, i'm assuming true for $i$ and trying to show that if $K_r=K_{r+1}= \dots = K_i \subset K_{i+1
}$ then this leads to a contradiction.

For $x \in K_{r+i}$ we have $T^{i-1}(x) \in K_{r+1}=K_r$, thus $x \in K_{r+i-1}$.
