# A decomposition of a finite-dimensional space

Suppose that $V$ is a finite-dimensional vector space over a field $k$ and suppose that $T: V \rightarrow V$ is a linear transformation. Set $V' = \cup_{n=1}^{\infty}\ker T^{n}$ and $V'' = \cap_{n=1}^{\infty}\text {Im }(T^{n})$. Show that $V$ is isomorphic to $V' \oplus V''$ via the canonical map $\phi: V' \oplus V'' \rightarrow V: \phi(v' \oplus v'') = v' + v''$.

Solution: We have the ascending chain which must stabilize as $V$ is finite-dimensional: $\{0\} \subset \ker T \subset \ker T^{2} \subset \cdots \subset V \implies n = \dim V \geq \dim \ker T^{k+1} \geq \cdots \geq \dim \ker T \geq 0$.

Similarly, we have a descending chain which must stabilize as well: $\{0\} \subset \cdots \subset \text{Im } (T^{k+1}) \subset \text{Im }(T^{k}) \subset \cdots \subset \text{Im }(T) \subset V$.

Thus there exists some $k$ such that $\ker T^{k} = \ker T^{k+j}$ and $\text{Im }(T^{k}) = \text{Im }(T^{k + j})$ for all $j\geq 1$.

We have $V' = \ker T^{k}$ and $V'' = \text {Im }(T^{k})$.

Let $W = \ker T^{k} \cap \text {Im }(T^{k})$. Then $w \in W \implies T^{k}x = w$ and $0= T^{k}w = T^{k +k}(x) = T^{k}(x) = w$. We have by the rank-nullity theorem that $\dim V = \dim(V' + V'')$ and it follows that $V$ is a direct sum of $V'$ and $V''$.

I am not sure if this is the way to go about this and I would like to see an argument that uses the map given in the problem.

The argument is mainly fine. Only "and $0= T^{k}w = T^{k +k}(x) = T^{k}(x) = w$" seems slightly incorrect. It is not necessarily true that $T^{k+k}(x)= T^k (x)$. (The space is stable but not necessarily each element.)
Instead I'd observe that as $\operatorname{im} T^k = \operatorname{im} T^{k+k}= T^k(\operatorname{im} T^k)$, the map $T^k$ restricted to $\operatorname{im} T^k$ is surjective and hence injective, which just means that $\operatorname{im} T^k$ contains only $0$ from $\ker T^k$.
Regarding using the map, you basically use it, as you assert that $V' + V" = V$.