Prove, that dimension of $W$ is equal the dimension of a minimal polynomial of $F$ Please help me in the following task
Let define $F: V \rightarrow V$ as an endomorphism on a finite-dimensional space $V$, and let $W = \mathrm{lin} (\mathrm{id}_v, F, F^2, F^3,\ldots) \subset \text{End}(V)$. Prove, that dimension of $W$ is equal the dimension of a minimal polynomial of $F$
 A: This is a basic fact about extensions. 
The elements of $W$ are of the form $b_{0} id_{V} + b_{1} F + \dots + b_{n} F^{n} = f(F)$, for $f(x) = b_{0} + b_{1} x + \dots + b_{n} x^{n}$ a polynomial. Divide $f(x)$ by the minimal polynomial $m(x)$ of $F$ to find
$$
f(x) = m(x) q(x) + r(x),\tag{div}
$$
where $r(x) = 0$, or $r(x)$ has degree less then $k$, the degree of $m(x)$. So evaluating (div) for $x = F$ we find
$$
f(F) = r(F) = a_{0} id_{V} + a_{1} F + \dots + a_{k-1} F^{k-1}
$$
for some $a_i$.
So $id_{v}, F, \dots , F^{k-1}$ are a set of generators for $W$. 
They are also linearly independent. One could show this by using the above argument and the uniquess of the remainder. However, one can also argue that if
$$
a_{0} id_{V} + a_{1} F + \dots + a_{k-1} F^{k-1} = 0\tag{ind}
$$
then $F$ is a root of the polynomial $g(x) = a_{0} + a_{1} x + \dots + a_{k-1} x^{k-1}$, which (is zero or) has degree less than the degree $k$ of the minimal polynomial. By the definition of the latter, $g(x) = 0$, so all $a_i = 0$ in (ind), which shows that $id_{v}, F, \dots , F^{k-1}$ are linearly independent.
