Infinite set of matrices span

Question

Assignment:

Let n be a positive integer.  Let A be an element of the vector space Mat(n,n,F), which has dimension n² over F.  Show that the span of the infinite set of matrices span(I_n, A, A², A³, ...) has dimension not exceeding n over F.

Defn of the linear space Mat(n,n,F):

The set of all n-by-n matrices with entries in F.  Mat(n,n,F ) is a vector space.  The additive identity in Mat(n,n,F ) is the n-by-n matrix all of whose entries equal zero.

Clues to solving the problem:

A is an element of Mat(n,n,F )

dim Mat(n,n,F ) = n²

I_n = Identity matrix.  1's on the diagonal and 0's else.

I_n, A, A², ... , A^{n^2}  -- n+1 matrices are linearly dependent.

Want to show that dim span(I_n, A, A², A³, ...), all infinitely many of them, are less than or equal to n.

Similar problem:

Assume that A is invertable (that is that A^-1 exists) then A^-1 is an element of span(I_n, A, A², A³, ...,A^n-1).

Reformulation: A^-1 = p(A) for a polynomial P

Proof:  Interpret A as M(T) for some T in L(Fⁿ).  let c be the minimum polynomial for T.

c = a₀ +a₁z+ ... + a_mz^m, m ≤ n

means that c(T) = 0,  assume a_m ≠ 0

a₀ +a₁T + ... + a_mT^m = 0 If a₀ = 0, then c(0)=0, and 0 is an eigenvalue for T.  T is not injective and A^-1 does not exist.  Since it DOES exist(due to the hypothesis), c(0)≠0, therefore a₀ ≠ 0

So, 1 = -(a₁/a₀)T - ... - (a_m/a₀)T^m

So, I_n = -(a₁/a₀)A - ... - (a_m/a₀)A^m

So, A^-1  = -(a₁/a₀) I_n - ... - (a_m/a₀)A^m-1

Provide complete and step by step solution for the question and show calculations and use formulas.