Jordan Canonical form of the operator $(T^2 - T)|_{\operatorname{Im}(T^4)}$.

Question

Suppose $T: V \to V$ is a nilpotent linear operator with $\dim (V) = 21$ and we know that

$$\dim (\ker(T)) = 6$$ $$\dim (\ker(T^2)) = 11$$ $$\dim (\ker(T^3)) = 15$$ $$\dim (\ker(T^4)) = 18$$ $$\dim (\ker(T^5)) = 20$$ $$\dim (\ker(T^6)) = 21$$

We have to find the Jordan Canonical form of the operator $(T^2 - T)|_{\operatorname{Im}(T^4)}$.

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What I have understood so far: $T$ is a nilpotent linear operator of order $6$. There will be $6$ Jordan blocks and size of the Jordan block we will get from the above information. Then how to proceed with the problem? Thank You!!

Answer 1

Most of the information is superfluous. All we need to know is that the dimension is $21$, that $T$ is nilpotent of order $6$ and that $\dim \ker T^4 =18$.

By the Rank Nullity Theorem the dimension of $U:=\text{Im}( T^4)$ is $3$.

Let $S=T(T-I)$; then $S^2 T^4=T^6 (T-I)^2=0$. So the minimal polynomial of $S|_U$ divides $X^2$. As $T^4\not=0$ the minimal polynomial of $S$ is not $1$. It cannot be $X$ either, else $T^5=-T^5(T-I)=0$ which is not so.

So the JCF of $S$ consists of a single $1\times 1$ block and a single $2\times 2$ block.