For any $n ≥ 2$, there exists a $n × n$ real matrix $A$ such that the set $\{A^p | p ≥ 1\}$ spans the $\Bbb R$-vector space $M_n(\Bbb R)$. True or false? 

For any $n ≥ 2$, there exists a $n × n$ real matrix $A$ such that the set $\{A^p \mid p ≥ 1\}$ spans the $\Bbb R$-vector space $M_n(\Bbb R)$.

I am not getting any clue on how to approach!! One thing that came in my mind is to deal with polynomials $f(x)=a_0+a_1x+a_2x^2+\cdots+$ which is satisfied by $A$ but this will not put any effect as we have to create an arbitrary matrix in the spanning set.
 A: This is false since Cayley Hamilton implies that $dimVect\{A^p, p\in\mathbb{N}\}\leq n+1$ and $dim(M_n(\mathbb{R})=n^2$.
A: FALSE!!!
By the Cayley-Hamilton theorem, $A$ satisfies the $n$-th degree polynomial
$\displaystyle \sum_0^n a_k A^k = (-1)^n \det(A - xI), \tag 1$
where
$a_n = 1; \tag 2$
this implies that
$A^n = -\displaystyle \sum_0^{n - 1} a_kA^k; \tag 3$
now consider that
$A^{n + 1} = AA^n = A \left (-\displaystyle \sum_0^{n - 1} a_kA^k \right ) = \displaystyle -\sum_0^{n - 1} a_k A^{k + 1} = -\sum_1^n a_{k - 1}A^k; \tag 4$
we thus see that $A^{n + 1}$ may be expressed in terms of the $A^k$, $1 \le k \le n$; now suppose there is some $m \in \Bbb N$ such that
$A^{n + m} = \displaystyle \sum_0^n b_kA^k; \tag 5$
then
$A^{n + (m + 1)} = AA^{n + m} = A\displaystyle \sum_0^n b_kA^k = \sum_0^n b_kA^{k + 1}$
$= b_nA^{n + 1} + \displaystyle \sum_0^{n - 1} b_kA^{k + 1} = -b_n \sum_1^n a_{k - 1}A^k + \sum_0^{n - 1} b_kA^{k + 1}; \tag 6$
taking (4) as the base case, we see that by virtue of (5) and (6) we have proved that every power of $A$ lies in
$\text{span} \{I, A, A^2, \ldots, A^n \}; \tag 7$
but
$\dim \text{span} \{I, A, A^2, \ldots, A^n \} \le n + 1, \tag 8$
whereas for $n \ge 2$
$\dim M_n(\Bbb R) = n^2 > n + 1; \tag 9$
thus 
$\text{span} \{I, A, A^2, \ldots, A^n \} \ne M_n(\Bbb R), \tag{10}$
as was to be proved.
