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I'm self-studying linear algebra and encountered a problem that I need help with. Here is the problem statement

Suppose $V$ is a finite-dimensional $\mathbb{C}$-vector space and $T\in\mathcal{L}(V)$ has $\lambda$ as an eigenvalue. Suppose also that $(z-\lambda)^m$ is a factor of the characteristic polynomial of $T$ and $(z-\lambda)^k$ is a factor of the minimal polynomial (and no larger powers of $z-\lambda$ in either case). If $\bar{v}$ is a Jordan basis for $T_{|G(\lambda,T)}$, find $\dim E(\lambda,T_{|G(\lambda,T)})$ and compute the corresponding matrix $M(T_{|G(\lambda,T)},\bar{v})$ when 1. $k=1$ and 2. $k=m$.

Notation wise, $E(\lambda,T)$ is the eigenspace of linear function $T$ with respect to eigenvalue $\lambda$ and $G(\lambda,T)$ the generalized eigenspace. $T_{|A}$ is the operator $T$ restricted to the set $A$ (i.e. the domain is set $A$).

I'm wondering if it is correct to state that $$ E(\lambda,T_{|G(\lambda,T)}) = E(\lambda,T) $$ as $E(\lambda,T) \subset G(\lambda,T)$. Also, in the case of $k=1$, I understand that the largest Jordan block is $1\times 1$. Am I correct in conjecturing that the corresponding matrix in this case is $\lambda I$? Any clue to solving this question is much appreciated :-)

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Let us find all eigenvectors of $E(\lambda, T_{|G(\lambda, T)})$. Clearly, $E(\lambda, T) \subset G(\lambda, T)$ is a set of eigenvectors, and by the definition of a generalized eigenvector, there are no more eigenvectors. Thus, $$E(\lambda, T_{|G(\lambda, T)}) = E(\lambda, T)$$


We can see that $T- \lambda\mathrm{id}$ is a nilpotent map on $G(\lambda, T)$, and $k$ is the index. So

$$k=1 \implies T-\lambda \mathrm{id} = \mathbf{0}$$

which implies that every vector is an eigenvector, or $G(\lambda, T) = E(\lambda, T)$. The corresponding matrix is $\lambda \mathrm{id}$. Now

$$k = m \implies (T-\lambda \mathrm{id})^m = \mathbf{0}, \: (T-\lambda \mathrm{id})^{m-1} \neq \mathbf{0}$$ Thus there exists basis vectors $v_1, \ldots, v_m$ so that $T - \lambda \mathrm{id}$ maps

$$v_1 \mapsto v_2 \mapsto \ldots v_m \mapsto \mathbf{0}$$ If $E(\lambda, T)$ is one dimensional, then the matrix looks like this:

$$ \begin{bmatrix} \lambda & 1 & 0 & \ldots & 0 & 0 \\ 0 & \lambda & 1 & \ldots & 0 & 0 \\ 0 & 0 & \lambda & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & \lambda & 1 \\ 0 & 0 & 0 & \ldots & 0 & \lambda \\ \end{bmatrix} $$

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