I am trying to prove the following:
Let $V$ be a finite-dimensional vector space. Consider an operator $T$ on $V$ such that $\text{dim range}(T)=\text{dim range}(T^2)$. Show that $V=\text{null}(T)\oplus \text{range}(T)$.
I have been given the following hint:
$\textit{Hint:}$ Show that there does not exist a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$.
I decided to try and prove the hint by contradiction…
Suppose that there exists a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$. Then
$$Ty\neq 0$$
and for every $v\in V$ we have $$Tv\neq y.$$
I have NO idea what to do with this information! I would greatly appreciate some hints/help! Thank you.
Best Answer
Because the range of $T$ contains the range of $T^2$, what you get is that the range of $T$ and of $T^2$ coincide. Because of the rank nullity formula, you also know the kernel of $T$ and $T^2$ coincide.
Pick an arbitrary vector $v\in V$. Because $Tv$ is in the image of $T$, there is some $w$ such that $Tv=T^2w$. This reads $T(v-Tw)=0$. It follows that you can write $$v=(v-Tw)+Tw\in \ker T+\operatorname{im}\, T$$ so it suffices to show the sum is direct.
Suppose then that $Tv=w$ and that $Tw=0$. Then $T^2v=0$, but because of the remark in the first paragraph, $v\in \ker T$ too, so that $Tv=w=0$.
For a personal opinion, I find the hint quite misleading -- it is completely unnecessary to argue by contradiction here, and might even obscure what is really going on. Note that you can really "chase the hypotheses" here to get a solution: you have a vector $v$ in $V$, and you want to write it as a direct sum of two things. You know something about the image of $T$ and $T^2$. It is natural to look at $Tv$. Now you can use what you know, and you get what you want, immediately. Similarly, to show the sum is direct you start with the information that $w\in \ker T\cap {\rm im}\, T$, and naturally a condition relating $\ker T$ and $\ker T^2$ pops up. But you know something about $\ker T$ and $\ker T^2$, and you can conclude.