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 $(T^2 – T)|_{\operatorname{Im}(T^4)}$.
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!!
Best Answer
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.