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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

$$ \begin{eqnarray*} \alpha_{00} &=& \sum_{k=0}^{\infty}a(0, 0, k) \\ &=& \sum_{k=0}^{\infty}a(0, 0, 2k + 1) \\ &=& \epsilon\sum_{k=0}^{\infty}a(1, 0, 2k) \\ &=& \epsilon\alpha_{10} \end{eqnarray*} $$


$$ \begin{eqnarray*} \alpha_{10} &=& \sum_{k=0}^{\infty}a(1,0,k) \\ &=& \sum_{k=0}^{\infty}a(1,0,2k) \\ &=& -\epsilon + \sum_{k=1}^{\infty}a(1,0,2k) \\ &=& -\epsilon - \epsilon\sum_{k=1}^{\infty}a(0,0,2k-1) \\ &=& -\epsilon\left(1 + \sum_{k=0}^{\infty}a(0,0,2k+1) \right) \\ &=& -\epsilon\left(1 + \alpha_{00}\right) \end{eqnarray*} $$

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.

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This special case can be done by inspection, but I cannot resist to mention that this is special case of von Neumann's result that $\sum_{k=0}^\infty A^k = (I-A)^{-1}$ (under some conditions on A). See e.g. Matrix analysis and applied linear algebra, Carl Dean Meyer, p. 527… – Martin Sleziak Mar 29 '11 at 7:14
@Martin. Right. The purpose of this exercise is to go the other way and find $(I - A)^{-1}$ by summing the series. It's somewhere in that wall of text ;) – knucklebumpler Mar 29 '11 at 7:19
@Martin. A nitpick. While John von Neumann did a lot of things, this time it wasn't him. It's Carl Gottfried Neumann's series (1877) and that precedes JvN's work by half a century. – t.b. Mar 29 '11 at 7:27
up vote 12 down vote accepted

Hint: Use the fact that $A^2 = -\epsilon^2 I$. (Then $A^3 = -\epsilon^2 A$, $A^4 = \epsilon^4 I$, etc.)

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That does the trick. Thanks. – knucklebumpler Mar 29 '11 at 7:39
You're welcome. While you're at it, you might want to try $\exp(A)$ (that is, $I+A+A^2/2! + A^3/3! + \dots$) too, just for fun. :) – Hans Lundmark Mar 29 '11 at 17:33

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) $$\sum A^n=\sum (-i\epsilon)^n=\frac{1}{1+i\epsilon}=\frac{1-i\epsilon}{1+\epsilon^2}= \frac{1}{1+\epsilon^2}\left(\begin{array}{cc}1&\epsilon\\-\epsilon&1\\\end{array}\right)$$

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