A non-homogenous linear recurrence: Why does my method fail? I'm trying to solve this recurrence relation: 
$$a_m = 8 \cdot a_{m-1} + 10^{m-1}, a_1=1$$
By the change of variable $\displaystyle b_m = \frac{a_m}{10^m}$ I obtained this linear non-homogenous relation:
$$b_m = \frac{8}{10}b_{m-1} + \frac{1}{10}, b_1 = \frac{1}{10}$$
Now I'm trying to solve this one. Since it's non-homogenous because of a constant, I thought maybe it was a good idea to use linear algebra by writing this recurrence relation in this matrix form:
$$\begin{bmatrix} b_m \\ b_{m-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{10} & \frac{8}{10}\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ b_{m-1} \end{bmatrix}$$
So, If we set $A=\begin{bmatrix} \frac{1}{10} & \frac{8}{10}\\ 0 & 1 \end{bmatrix}$, my recurrence relation can be written like this:
$$\begin{bmatrix} b_m \\ b_{m-1} \end{bmatrix} = A^m \begin{bmatrix} 1 \\ a_1 \end{bmatrix} = A^m \begin{bmatrix} 1 \\ \frac{1}{10} \end{bmatrix}$$
The problem arises when I want to calculate the eigenvalues and eigenvectors of $A$ to find $A^m$. It involves square roots and stuff and apparently they don't get canceled. Another problem is that if I write the same recurrence relation in the following way my calculations TOTALLY changes, I mean eigenvalues change, so, eigenvectors will change too, and it seems like a totally different problem:
$$\begin{bmatrix} b_m \\ b_{m-1} \end{bmatrix} = \begin{bmatrix} \frac{8}{10} & \frac{1}{10}\\ 1 & 0 \end{bmatrix} \begin{bmatrix} b_{m-1} \\ 1 \end{bmatrix}$$
Using wolfram alpha I know the exact form for $b_m$:
$$b_m = \frac{1}{2} ( 1 - (\frac{4}{5})^m ) $$
Which agrees with what I expect.
Is there something wrong with my method that such complications arise?

Your method fails because you can't exponentiate $A$ as you did; the output of one step is not the input to the next one.  You want instead something of this form:$$\begin{bmatrix} b_m \\ b_{m-1} \end{bmatrix}=A\begin{bmatrix} b_{m-1} \\ b_{m-2} \end{bmatrix}$$
