Describe the Jordan form of a linear operator $T:\mathbb{R}^7\to\mathbb{R}^7$ with characteristic polynomial $$p_T(t)=(t-1)^2(t-2)^4(t-3)$$ and such as $\dim(\ker(T-2I))=2, \dim(\ker(T-I))=1$ and $\ker(T-2I)^3\neq\ker(T-2I)^2$
My attempt:
As we have the characteristic polynomial and the dimension of the subspespaces associated to the eigenvalues, then I can say that the Jordan form to the operator is
\begin{bmatrix} 1 & & & & & & \\ 1 & 1 & & & & & \\ & & 2 & & & & \\ & & 1 & 2 & & & \\ & & & & 2 & & \\ & & & & 1 & 2 & \\ & & & & & & 3 \\ \end{bmatrix}
Is this correct?
I don't understand the condition $\ker(T-2I)^3\neq\ker(T-2I)^2$, what does it means and what is the purpose of that? Thanks in advance!
The Jordan blocks associated with eigenvalues $1$ and $3$ are correct. To deduce the correct Jordan block associated with eigenvalue $2$, first note that $\dim(\ker(T - 2I)) = 2$ implies that there are two possibilities below:
\begin{align*} & \text{(a).}\; J_a = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 2 \end{bmatrix}, \quad \text{(b).}\; J_b = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}. \end{align*} It is easy to verify that \begin{align*} (J_a - 2I_{(4)})^2 = (J_a - 2I_{(4)})^3 = 0, \end{align*} while \begin{align*} (J_b - 2I_{(4)})^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \neq 0 = (J_b - 2I_{(4)})^3. \end{align*}
Therefore, case (a) contradicts with the condition $\ker((T - 2I)^2) \neq \ker((T - 2I)^3)$, implying that case (b) is the unique compatible Jordan block.