Let $M \in M_{4\times 4}(\mathbb{C})$ and $W$ be a subspace of $M_{4\times 4}(\mathbb{C})$ such that $W$=$L({M^n: n\geq 0})$ .What will be the possibilities for the dimension of $W$ ?
Attempt:
I found $W$=$L(I,M,M^2,M^3,M^4,...,M^n,...)$ and $L(I,M,M^2,M^3,M^4,...,M^n,...)$ is not linearly independent set as matrices $M^5 ,M^6,...$ can be written as a linear combination of $I,M,M^2,M^3,M^4$. So dimension of $W$ will be $\leq 4$ .
Case-$1$ > $\operatorname{dim}W=4$ ,when $M_{4\times4}$ is not nilpotent matrix .
Case-$2$> $\operatorname{dim}W=3$ ,when $M_{4\times4}$ is nilpotent matrix of index $4$ .
Case-$3$> $\operatorname{dim}W=2$ ,when $M_{4\times4}$ is nilpotent matrix of index $3$ .
Case-$4$> $\operatorname{dim}W=1$ ,when $M_{4\times4}$ is nilpotent matrix of index $2$ .
Case-$5$> $\operatorname{dim}W=0$ ,when $M_{4\times4}$ is null matrix .
Question:
Is my method correct? or something which is missed ?
Edits:
$(1)$
here, $L$ represents the linear span set.
$(2)$
Case-$5$ will not exist so, $\operatorname{dim}W\neq 0$ .
Your method is correct, except that the subspace always contains $I$ as per the statement of the problem, so the dimension cannot be $0$. It can be $1,2,3,4$ as you have shown.
It is, however, possibly much easier to see it on diagonal matrices, because, for $p\in\mathbb C[x]$ and $D=\begin{bmatrix}d_1&0&0&\cdots&0\\0&d_2&0&\cdots&0\\0&0&d_3&\cdots&0\\\cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&\cdots&d_n\end{bmatrix}$, we have $p(D)=\begin{bmatrix}p(d_1)&0&0&\cdots&0\\0&p(d_2)&0&\cdots&0\\0&0&p(d_3)&\cdots&0\\\cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&\cdots&p(d_n)\end{bmatrix}$, so the minimal polynomial of $D$ looks like this: $\mu_D(\lambda)=\prod_{i\in I}(\lambda-d_i)$ where $I$ is the set of indices that gives you all the mutually different $d_i$'s.
Now the subspace you are talking about here is of dimension $\deg\mu_D$, so you can easily find matrices with minimal polynomial of degree $1,2,3,4$, for example: $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$ (all diagonal components the same), $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&2\end{bmatrix}$ (two different components), $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&3\end{bmatrix}$ (three different) and $\begin{bmatrix}1&0&0&0\\0&2&0&0\\0&0&3&0\\0&0&0&4\end{bmatrix}$ (all four different).