TITLE: Summing Matrix Series
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: if you dont want to use the geometric series directly on the matrix space, you can also think of this as the complex number $-i\epsilon$ under the correspondence $$a+bi\to\left(\begin{array}{cc}a&-b\\b&a\\\end{array}\right)$$ and use the geometric series (for complex numbers of norm less than 1) $$\sum A^n=\sum (-i\epsilon)^n=\frac{1}{1+i\epsilon}=\frac{1-i\epsilon}{1+\epsilon^2}=
\frac{1}{1+\epsilon^2}\left(\begin{array}{cc}1&\epsilon\\-\epsilon&1\\\end{array}\right)$$