Linear Algebra - Kernel and Linear Transformation Let $V $ be an n-dimensinal complex vector space, and let $\ T: V \rightarrow V $ be a linear transformation.
Given that $$K_i = Ker \ T^i$$
Have shown that $ K_i \subseteq K_{i+1} for \ each\ i $ and that there exists a non-negative interger r such that $ K_r = K_{r+1}$. Struggling to show that $K_r=K_{k+j} \ \forall j \ge 1$ and that $ V = K_r \oplus T^r(V)$.
 A: We prove that 
$$K_{r+j}=K_{r+j+1}\quad \forall j\ge0$$
using this: If $x\in K_{r+j+1}$ then $f^j(x)\in K_{r+1}=K_r$ so $x\in K_{r+j}$. 
moreover if $y\in T^r(V)\cap K_r$ then $y=T^r(x)$ and $T^r(y)=T^{2r}(x)=0$ so $x\in K_{2r}=K_r$ hence $y=T^r(x)=0$.
The desired result follows by using the Rank nullity theorem.
A: Some Hints:
If $T^{r+2}(x)=0,$ then $T(x)\in K_{r+1}=K_{r},$ so $x\in K_{r+1}=K_r$. Then you can show the first point by induction.
As to the second point, if $x\in K_r\cap T^r(V),$ then $x=T^r(y)$ for some $y\in V$, hence $y\in K_{2r}=K_r$, i.e. $x=0$. Finally, $V=T^r(V)\oplus K_r$ for dimensional reasons, i.e. the dimension-theorem.
Hope this helps.
A: If $v \in \ker T^{r+2}$, then $T^{r+2}v = 0$. 
Which is the same as saying that $T^{r+1}Tv = 0 \ \Longleftrightarrow Tv \in \ker T^{r+1} = \ker T^r$.
As for the equality
$$
V = \ker T^r \oplus \mathbb{im}\ T^r \ ,
$$
try to use the fact that, in general, for any endomorphism $f: V \longrightarrow V$ you have that $\mathbb{dim}\ker f + \mathbb{dim}\ \mathbb{im}\ f = \mathbb{dim}\ V$, together with the computation of $\ker T^r \cap \mathbb{im}\ T^r$.
