Compute $A^n$ where $A^2+bA+cI=0$ Let $A$ be a complex matrix such that $$A^2+bA+cI=0,$$
where $I$ is the identity matrix and $b,c\in \mathbb{C}$. I am interested in finding a formula for $A^n$ in terms of $A$ and $I$. The binomial formula  is not giving an answer I think. Maybe using $A^2=-bA-cI$, then $A^3=-bA^2-cA$, etc. But the computation becomes so complicated to find a formula by induction.

You have $$ A^{n+2} = b_nA + c_nI $$
where $b_0 = -b, c_0 = -c$
Multiplying both sides by $A$ gives
$$ A^{n+3} = b_nA^2 + c_n A = b_n(-bA-cI)+c_nA = (c_n-bb_n)A - cb_nI $$
therefore
$$ b_{n+1} = -bb_n + c_n, \ c_{n+1} = -cb_n $$
You may find an implicit relation by solving the linear system
$$ \left(\matrix{b_n\\c_n}\right)=M^n\left(\matrix{-b\\-c}\right) $$
where
$$ M = \left(\matrix{-b&c\\-c&0}\right) $$
then we have the characteristic
$$ \lambda^2 + b\lambda + c^2 = 0 $$
The two eigenvalues are $\lambda_1,\lambda_2$, where $b = -(\lambda_1+\lambda_2), c^2 = \lambda_1\lambda_2$
The eigenvectors are
$$ \vec{v_1} = \left(\matrix{-\lambda_1\\c}\right), \ 
\vec{v_2} = \left(\matrix{-\lambda_2\\c}\right) $$
Thus
$$ M^n\left(\matrix{-b\\-c}\right) = -M^n(\vec{v_1}+\vec{v_2}) = -{\lambda_1}^n\ \vec{v_1} - {\lambda_2}^n\ \vec{v_2} $$

If $\lambda_1 = \lambda_2$ then we have the generalized eigenvectors
$$ \vec{v_1} = \lambda \left(\matrix{-1\\1}\right), \ \vec{v_2} = \left(\matrix{-2\\1}\right) $$
Then 
$$ M^n\left(\matrix{2\lambda\\-\lambda}\right) = -M^n \lambda \vec{v_2} = \lambda^{n+1}\vec{v_2} + n\lambda^n \vec{v_1} $$
