Multiple choice question about the dimension of a space of $10 \times 10$ complex matrices 
Problem:     Let $A \in M_{10}(\Bbb C)$ ,the vector space of $10 \times 10$ matrices with entries in $\Bbb C$.
Let $W_A$ be subspace of  $M_{10}(C)$ spanned by $\{\,A^n : n\geq 0\,\}$
Then which of the following correct ?
1) $\dim(W_A) \leq 10$ 
2) $\dim(W_A) < 10$ 
3) $10 <\dim(W_A)<100$ 
4)$\dim(W_A)=100$ 
Solution:I think $\dim(M_{10}(\Bbb C))=100$
So $\dim(W_A) \leq 100$ 
elements of  $W_A$ is linear combination of $\{\,A^n : n\geq 0\,\}$

After that I have no idea.
 A: Note, that this will depend on $A$, for example, if $A = \def\Id{\mathop{\rm Id}}\Id$, we have $W_A = \mathbb C\cdot \Id$, so $\dim W_A = 1$, if $A$ is not a multiple of $\Id$, we have $\{\Id, A\} \subseteq W_A$ is linear independent, hence $\dim W_A \ge 2$. 
In general, if $\mu_A$ denotes the minimal polynomial of $A$, that is the uniquely determinend smallest degree normed member of 
$$ I_A = \{p \in \mathbb C[X] \mid p(A) = 0\} $$
($I_A$ is an ideal in $\mathbb C[X]$ and $\mu_A$ is its normed generator). Then we have $\mu_A(A) = 0$, so $A^{\deg \mu_A}$ is a linear combination of $\{A^i\mid i < \deg\mu_A\}$ and this cannot happen for lower powers (since this would give a lower degree member of $I_A$). Hence $\dim W_A = \deg \mu_A$.

Addendum (after the "options" were added in the OP): Note that we know by Cayley-Hamilton, that $\chi_A(A) = 0$, where $\chi_A(X) = \det(X\Id - A)$ denotes $A$'s characteristic polynomial. So $\dim W_A \le \deg \chi_A = 10$. This tells us that options 3) and 4) are wrong. To show that 1) is the right answer, we will give a matrix such that $A^i$ for $i = 0,\ldots, 9$ are linear independent, hence $\dim W_A = 10$ can occur. Let $A = (a_{ij})$ where 
$$ a_{ij} = \begin{cases} 1 & j = i+1\\ 0 & \text{otherwise} \end{cases}
$$
Then $A^k = (a_{k,ij})_{ij}$ where
$$
  a_{k,ij} = \begin{cases} 1 & j = i+k\\ 0 &\text{otherwise} \end{cases} 
$$
Now for any $\lambda_k \in \mathbb C$ we have 
$$ \left(\sum_{k=0}^{9} \lambda_k A^k\right)_{ij} = \begin{cases} a_{j-i} & j-i \in [0,9]\\ 0 & \text{otherwise}\end{cases} $$
which gives linear independence.
A: Hint: Think of Characteristics Polynomial. Suppose your space is of $n\times n$ matrix.
So there exists $c_0,c_1,\dots,c_n$ not all $0$ such that $c_0 I+c_1A+\dots+c_nA^n=0$  and 
so  in your case $\{A^0,A,A^2,\dots,A^{10}\}$ is linearly dependent. can you take from here?So dimension will be $\le 10$
