# 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.

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The part $K_r \subset K_{r+i}$ should be easy. Here's a hint for the converse: let $v \in K_{r+2}.$ Then $T^{r+2}(v) = 0 \rightarrow T(v) \in K_{r+1} = K_r$ (by hypothesis) $\rightarrow v \in K_{r+1} \rightarrow K_{r+2} \subset K_{r+1} \rightarrow K_{r+2} = K_r.$ –  Gerben Oct 31 '11 at 15:19

You've been given good hints on showing that $K_r = K_{r+i}$ for $i\geq 0$.

To show that $V=K_r\oplus T^{r}(V)$, note that the Rank-Nullity Theorem guarantees that $\dim(K_r) + \dim(T^r(V)) = \dim(V)$, so you really only need to show that the sum is direct; that is, that $K_r\cap T^{r}(V)=\{\mathbf{0}\}$.

So, assume that $\mathbf{v}\in K_r\cap T^{r}(V)$. You want to show that $\mathbf{v}=\mathbf{0}$. Since $\mathbf{v}\in T^r(V)$, there exists $\mathbf{w}\in V$ such that $T^r(\mathbf{w})=\mathbf{v}$; since $\mathbf{v}\in K_r$, this means that $\mathbf{0}=T^r(\mathbf{v})=T^r(T^r(\mathbf{w}))=T^{r+r}(\mathbf{w})$, so $\mathbf{w}\in K_{r+r}$. But since $K_r=K_{r+r}$...

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I was so almost there! thank you, that was getting fustrating. –  Freeman Oct 31 '11 at 16:42

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

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So if we have assumed for $i-1$ and $x\in K_{r+i}$ and we know $T^{i-1}(x)\in K_{r+1}$ as $T^{r+1}T^{i-1}(x)=T^{r+i}(x)$ as we have assumed for $i-1$. Therefore as you say $T^{i-1}(x)\in K_{r}$ $\Rightarrow K_{r+i} \subseteq K_{r+i-1}$ which contradicts the fact $K_{r+i-1} \subset K_{r+i}$ –  Freeman Oct 31 '11 at 15:47
@LHS: If you have both $K_{r+i}\subseteq K_{r+i-1}$ and $K_{r+i-1}\subseteq K_{r+i}$ then it means $K_{r+i}=K_{r+i-1}$. –  Joe Johnson 126 Oct 31 '11 at 15:57
No proofs of contradiction here. –  Martin Brandenburg Nov 1 '11 at 10:59

HINT: $T^{-1}(K_{r+i})=K_{r+i+1}$ for all $i\geq 0$. Also use your known result $K_r=K_{r+1}$.

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