I need to sum the series
$I + A + A^2 + \ldots$
for the matrix
$A = \left(\begin{array}{rr} 0 & \epsilon \\ -\epsilon & 0 \end{array}\right)$
and $\epsilon$ small. The goal is to invert the matrix $I - A$. The text says to use a geometric series but I had a hard time finding it. I'm studying on my own so I can't ask my teacher. The way I did it follows. I know it isn't quite rigorous (I assume the series in question converge) so I'd like to see how I'm supposed to do it.
We see that
$\left(\begin{array}{rr} 0 & \epsilon \\ -\epsilon & 0 \end{array}\right) \left(\begin{array}{rr} a_{00} & a_{01} \\ a_{10} & a_{11} \end{array}\right) = \left(\begin{array}{rr} \epsilon a_{10} & \epsilon a_{11} \\ -\epsilon a_{00} & \epsilon a_{01} \end{array}\right)$
so if we let $a(i, j, k)$ be entry $a_{ij}$ in the $k$'th power of $A$ then we see that
$ a(0, 0, k) = \epsilon a(1, 0, k-1) $
$ a(1, 0, k) = -\epsilon a(0, 0, k-1) $
Then, letting $\alpha$'s denote the entries in the sum without $I$ added in, we see that
$ \begin{eqnarray*} \alpha_{00} &=& \sum_{k=0}^{\infty}a(0, 0, k) \\ &=& \sum_{k=0}^{\infty}a(0, 0, 2k + 1) \\ &=& \epsilon\sum_{k=0}^{\infty}a(1, 0, 2k) \\ &=& \epsilon\alpha_{10} \end{eqnarray*} $
and
$ \begin{eqnarray*} \alpha_{10} &=& \sum_{k=0}^{\infty}a(1,0,k) \\ &=& \sum_{k=0}^{\infty}a(1,0,2k) \\ &=& -\epsilon + \sum_{k=1}^{\infty}a(1,0,2k) \\ &=& -\epsilon - \epsilon\sum_{k=1}^{\infty}a(0,0,2k-1) \\ &=& -\epsilon\left(1 + \sum_{k=0}^{\infty}a(0,0,2k+1) \right) \\ &=& -\epsilon\left(1 + \alpha_{00}\right) \end{eqnarray*} $
so $\alpha_{00} = \epsilon\alpha_{10}$ and $\alpha_{10} = -\epsilon(1 + \alpha_{00})$ which we can solve for the $\alpha$'s.
It's pretty much the same for the other two. I feel like there has got to be a better way to do this.