$V=N(T^m)\oplus R(T^m)$

Question

I was reading this post to show that if $T : V\to V$ is a linear map between finite dimensional vector space, then there exists some $k$ such that $V=N(T^m)\oplus R(T^m)$ but I don't understand why if $N(T^m)\cap R(T^m)={0}$ every element $z\in V$ can be written as a sum of an element in the nul space and the range space.

I can write $z=(z-T^k(z))+T^k(z)$

So than $T^k(z-T^k(z))=T^k(z)-T^{2k}(z)$ but is this $0$?

Answer 1

If $\operatorname{Ker}(T^m) \cap \operatorname{Im}(T^m) = \{0\}$, then \begin{align*}\dim(\operatorname{Ker} (T^m) + \operatorname{Im} (T^m)) &= \dim \operatorname{Ker} (T^m) + \dim \operatorname{Im} (T^m) - \dim(\operatorname{Ker} (T^m) \cap \operatorname{Im} (T^m)) \\ &= \dim \operatorname{Ker} (T^m) + \dim \operatorname{Im} (T^m) = \dim V, \end{align*} by the rank-nullity theorem. Thus, $\operatorname{Ker} (T^m) + \operatorname{Im} (T^m)$ is a subspace of $V$ of full dimension, and hence is equal to $V$. The sum is direct because the subspaces intersect trivially.

As for what you wrote, while it will be true that any $z$ can be written uniquely as a sum of elements of $\operatorname{Ker} (T^m)$ and $\operatorname{Im} (T^m)$, i.e. there is some $w \in V$ such that $$z = (z - T^mw) + T^mw$$ where $z - T^mw \in \operatorname{Ker}(T^m)$, there's no guarantee that $w = z$.

Answer 2

You know that by Rank-Nulity theorem- $$dim V= dim(N(T^k))+dim(R(T^k))$$ We also know that for any two subspaces (say $S$ & $W$) of a finite dimensional vector space $ V$ $$dim(S+W)=dim(S)+dim(W)-dim(S\cap W)$$ Here they made use of that fact $$dim(N(T^k)+R(T^k))=dim(N(T^k))+dim(R(T^k))-dim(N(T^k)\cap R(T^k))$$ Now we have $dim(N(T^k)\cap R(T^k))=0$

Hence$$dim(N(T^k)+R(T^k))=dim(N(T^k))+dim(R(T^k))=dim V$$$$\implies V=N(T^k)+R(T^k)$$ and therefore $V=N(T^k)\oplus R(T^k)$.

The thing you should note here is the following- The way in which you are trying to decompose the elements is not the way in which the elements are decomposed in V