# Jordan form of linear operator

Given the linear operator $$T: \mathbb{R}^4 \rightarrow \mathbb{R}^4: T^3=-4T^2$$ and $$\dim \text{Im}\ T=\dim\ker T$$.
Prove that exists Jordan Form of $$T$$ in $$\mathbb{R}$$ and find all of the possible Jordan Forms of $$T$$.
I'm sorry for asking without providing my own thoughts, but I don't have any ideas where to start from yet. Any help is very appreciated. Thank you in advance.

By the assumption on the dimension and the rank theorem, the kernel and image are both 2-dimensional. The fact that $$T^3 = -4T^2$$ means that the minimum polynomial $$p(t)$$ of this operator must divide $$t^2(t+4)$$. Checking the possibilities, we see that:
1. $$p(t) = t$$ would imply $$T \equiv 0$$, which is not true.
2. $$p(t) = t^2$$ implies that there is a Jordan block of size 2 corresponding to the eigenvalue 0. Since the kernel has size 2, we need exactly 2 blocks corresponding to this eigenvalue.
3. $$p(t) = t+4$$ would imply $$T = -4I$$ which is not true.
4. $$p(t) = t(t+4)$$ means the operator is diagonalisable (since the minimum polynomial splits) with 0 or -4 along the diagonal, so it must be $$\operatorname{diag}(0, 0, -4, -4)$$ upto ordering.
5. $$p(t) = t^2(t+4)$$: in this case there is a Jordan block of size 2 corresponding to the eigenvalue 0, and 2 Jordan blocks of size 1 corresponding to the eigenvalue -4.
$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0\\ 0 & 0 & 0 & -4 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0\\ 0 & 0 & 0 & -4 \end{bmatrix}$$ upto permutation of the blocks.