TITLE: Linear operator T on $ \mathbb{R}^5$. $T^4 \neq 0$, but $T^5 = 0$. What's the rank of $T^2$?
QUESTION [5 upvotes]: The problem:

Linear operator T  on $ \mathbb{R}^5$. $T^4 \neq 0$, but $T^5 = 0$. What's the rank of $T^2$?

I am learning linear algebra and preparing for a test.
For this problem, I am able to show that $range(T^4) \subseteq nullspace(T)$ since there exists $T^4x = y$, and $T^5x = Ty = 0$. $y$ is arbitrary and can represent the range of $T^4$. Then I am stuck.
I understand this is a duplicated question but I don't know Jordan canonical form or eigenspaces yet. So I couldn't understand the answer there. Could you help with it using only linear transformation theorems?

REPLY [3 votes]: If
$$
T(\Bbb R^5)=\Bbb R^5\;,
$$
this would imply $\ker T=0$ (from the well known formula $\dim\ker T+\dim\operatorname{Im}T=\dim\Bbb R^5=5$) and the OP pointed out that $\{0\}\neq T^4(\Bbb R^5)\leq\ker T$.
Thus
$$
T^j(\Bbb R^5)=T^{j-1}T(\Bbb R^5)\lneq T^{j-1}(\Bbb R^5)
$$
for every $j=2,\dots,5$.
So we get the following strict chain
$$
\{0\}=T^5(\Bbb R^5)\lneq T^4(\Bbb R^5)\lneq T^3(\Bbb R^5)\lneq T^2(\Bbb R^5)\lneq T(\Bbb R^5)\lneq\Bbb R^5
$$
from which it immediately follows that
$$
\operatorname{rk} T^2=3\;.
$$