Find the Dimension of $W=\operatorname{span}\{I,A,A^2,A^3,\ldots,A^m,\ldots\}$ 
Let $A \in M^{\Bbb C}_{n \times n}$ be a square matrix with a minimal polynomial of the degree $k$.
Find the dimension of  $W=\operatorname{span} \{ I, A, A^2, A^3, \ldots, A^m, \ldots\}$
The question assumes finite dimension.

I think I have to use induction here.
Thanks,
Alan
 A: If $m(x)=\sum_{i=0}^k a_ix^i$ is the minimal, then $m(A)=0$. Now, I affirm that $A^{k}\in \text{span}\{I, ..., A^{k-1}\}$, because $A^k=-1/a_k \sum_{i=0}^{k-1} a_iA^i$
A: Here is an example which will hopefully function as a hint.  Suppose that $A^2+3A+6I=0$, so that $A^2=-3A-6I$.  Then $A^3=A(A^2)=-3A^2-6A=-3(-3A-6I)-6A$.  Similarly, we multiplying this by $A$ and rewriting, we could express $A^4$ as a linear combination of $A$ and $I$.  Now, use induction.  Now, generalize.  This should hint you at what the answer is (and prove one direction of it).  
A: You don't actually need finite dimension of the vector space acted upon, the existence of a minimal polynomial for the linear operator (I'll call it $T$) suffices (but of course finite dimension does imply this existence, whereas in infinite dimension no minimal polynomial is guaranteed to exist).
Having a linear relation $\sum_{i=0}^dc_iT^d=0$ is precisely saying that the polynomial $\sum_{i=0}^dc_iX^d$ annihilates$~T$, so the fact that $T$ has a minimal polynomial of degree$~k$ means that the family $[T^0,\ldots,T^{k-1}]$ is linearly independent, while the family $[T^0,\ldots,T^k]$ is linearly dependent, which must be because $T^k\in W$ where $\def\Span{\operatorname{Span}}W=\Span(T^0,\ldots,T^{k-1})$. The former means that $\dim W=k$, while the latter means the subspace $W$ is $T$-stable (because $T$ sends each of the generators to a vector of$~W$). But then clearly $T^m\in W$ for all$~m\in\Bbb N$, so that $W=\Span(T^0,\ldots,T^k,\ldots)$, and we are done.
A: $\text{dim}W=k$ (well, if k>0).
Since the minimal polynomial of $A$ has the degree k, it means that there exist $a_i,i=0...k$:
$$
\sum_{i=0}^ka_iA^i=0.
$$
Moreover, $a_k\neq{0}$.
Can you now prove yourself that there exists a linear combination of any $k+1$ powers of $A$ equal to zero?
Since the polynomial is minimal, no lower-degree linear dependency is possible.
