Determine if a statement can haapen let $ V= M_{10}\left(F\right) $ a vector space over some field $ \mathbb{F} $
I have to determine if its possible that exists $ A\in M_{10}\left(F\right) $
such that $ M_{10}\left(F\right)=span\left\{ A^{i}:0\leq i\leq100\right\}  $
I guess it cant be. But I cant prove why. I tried to separate
the cases - if $ A $ is invertible and if it is not. But still, cant find anything that would help me decide.
I noticed that there are 101 matrices in the set $ \left\{ A^{i}:0\leq i\leq100\right\}  $  so one of them has to be linear independent in the rest of them. But I cant see how to continue
Thanks in advance
 A: The answer is indeed no. This is trivial if we can use the Cayley Hamilton theorem.
Note that each element of the span of $\{A^i : 0 \leq i \leq 100\}$ can be written in the form $p(A)$ for some polynomial $p$. However, the characteristic polynomial $\chi(x)$ (which has degree $n = 10$) satisfies $\chi(A) = 0$.
Note that any polynomial $p(x)$ can be written in the form
$$
p(x) = q(x) \chi(x) + r(x)
$$
with $r$ of degree at most $n-1$ via polynomial division. This means that
$$
p(A) = q(A)\chi(A) + r(A) = r(A).
$$
Because $r$ has degree at most $n-1$, $p(A) = r(A)$ is a linear combination of the elements $I,A^1,\dots,A^{n-1}$.
So, the span of $\{A^i : 0 \leq i \leq 100\}$ is at most $n$-dimensional, which means it cannot contain all of $V$.

Without Cayley-Hamilton:
Case 1: $A$ is a multiple of $I$; in this case it is clear that the powers of $A$ do not span $V$.
Case 2: $A$ is not a multiple of $I$. In this case, there exists at least one matrix $B$ for which $AB \neq BA$. It follows that $B$ is not an element of the span of $\left\{ A^{i}:0\leq i\leq100\right\}$.
