Let V be a finite-dimensional vector space, and let $T:V \rightarrow V$ be linear.
Prove that $V=R(T^k) \oplus N(T^k)$ for some positive integer k.
For this problem, I just can't get the gist of it. I know that we can extend $V=R(T^k) \oplus N(T^k)$ to Rank-nullity theorem, and by that theorem we can see that rank is finite dimensional and is within a certain range. Can someone tell the general idea of how the proof works for this problem?
Best Answer
Hints:
The direct sum $V=R(T^k)\oplus N(T^k)$ means that every element of $V$ can be written as a sum of elements from $R(T^k)$ and from $N(T^k)$, and that the intersection $N(T^k)\cap R(T^k)$ is $=\left\{0\right\}$.
Let's leave it as an exercise that if $N(T^k)\cap R(T^k)=\left\{0\right\}$, then the first condition holds automatically.
Now note that there is an integer $k$ such that $N(T^k)=N(T^{2k})$ (why?). Fix this $k$. Then take $x\in N(T^k)\cap R(T^k)$. Since $x\in R(T^k)$, there is $y\in V$ such that $x=T^k(y)$. Since $x\in N(T^k)$, we see $T^{2k}(y)=0$. Since $N(T^k)=N(T^{2k})$, it follows that $x=T^k(y)=0$. Therefore $N(T^k)\cap R(T^k)=\left\{0\right\}$.
Hope this helps.