Linear algebra proof by induction using span. $A \in M_n(R)$ and I is the n x n identity matrix.
Let $S$ = span $\{ A,I\}$ in the vector space $M_n(R)$.
If $A^2 \in S$, prove by induction that $A^k \in S$ for all positive integers $k$.
What I have done:
I can show that $A^2 = c_1A + c_2I$
And that $A^3 = AA^2 = A(c_1A + c_2I) = (c_1^2 + c_2)A + (c_1 c_2)I$
So, $A^3 \in S$.
And I can then show that $A^{k+1}$ can be written in the same format.
The problem is that I'm using $A^3$ as my base case.
How can I do this without using $A^3$ as the base case?
I think it has something to do with the span of a set of vectors.

The base case is, as usual, $k=0$, which is true because $A^0=I\in W$.
Suppose the statement holds for $k$. Then $A^k=\alpha A+\beta I$ and so
$$
A^{k+1}=AA^k=A(\alpha A+\beta I)=\alpha A^2+\beta A\in W
$$
because $A^2,A\in W$ and $W$ is a subspace.
