Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$ Let 
$$A=
        \begin{bmatrix}
        -1 & 6 & -9 \\
        -11 & 24 & -33 \\
        -6 & 12 & -16 \\
        \end{bmatrix}
$$
a) Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$
b) Write $A^n$ in a form of $a_nA+b_nI$

If matrix $A^4\in Span\{A,I\}$ then $A^4=\alpha A+ \beta B$.
I had an idea to find eigenvalues and eigenvectors of $A$, so I can diagonalize the matrix and say that $A^4=SD^4S^{-1}$. However, I am not sure is that done without calculating $A^4$.
Thank you all in advance.
 A: Notice that $A^2=5A-6I$.
Hence for any $n>0$, $A^n \in Span \left\{ A,I\right\}.$
To see this, for example, to compute $A^3$, we have
$A^3=5A^2-6A=5(5A-6I)-6A$.
A: Use this $A^2=5A-6I$ and calculate $A^4$. $$A^4=65A-114I$$
A: The characteristic polynomial of $A$ is
$$ p_A(\lambda) = \det(\lambda I - A) = \lambda^3 - 7\lambda^2 + 16\lambda - 12 = (\lambda - 3)(\lambda - 2)^2. $$
Since part (b) asks you to show in particular that $A^2 \in \operatorname{span} \{ I, A \}$, the minimal polynomial of $A$ must be of degree at most two. Since the minimal polynomial of $A$ must have the same linear factors as the characteristic polynomial, we then must have $m_A(\lambda) = (\lambda - 3)(\lambda - 2) = \lambda^2 - 5\lambda + 6$. This indeed holds:
$$ (A - 3I)(A - 2I) = \begin{pmatrix} -4 & 6 & -9 \\ -11 & 21 & -33 \\ -6 & 12 & -19 \end{pmatrix} \begin{pmatrix} -3 & 6 & -9 \\ -11 & 22 & -33 \\ -6 & 12 & -18 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
(you can deduce this without calculating by noticing that $1 = \operatorname{rank}(A - 2I) = 3 - \dim \ker(A - 2I)$ and so $A$ is diagonalizable and the minimal polynomial must be $(\lambda - 3)(\lambda - 2)$).
Thus, $A^2 = 5A - 6I$ and at this point you know that $A^n \in \operatorname{span} \{ I, A \}$ for all $n \geq 0$ (and in particular $n = 4$) without explicitly calculating $A^4$ or the coefficients. Finally, if we write $A^n = a_n A + b_n I$ we must have
$$ A^{n + 1} = a_{n+1} A + b_{n+1}I = A(A^n) = A(a_n A + b_n I) = a_n A^2 + b_n A = a_n (5A - 6I) + b_n A = (5a_n + b_n) A - (6a_n) I. $$
Since $A,I$ are linearly independent we get $a_{n+1} = 5a_n + b_n, b_{n+1} = -6a_n$ which is a recursion relation that can be solved explicitly together with the initial conditions $a_1 = 5, b_1 = -6$. 
