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Consider a symmetric matrix $A$ with non-negative integer coefficients. It appears that the geometric series $\sum_{i \geq 0}A^i$ will converge to a matrix $B$ if the spectral radius (the largest eigenvalue in absolute value) is less than 1.

I would like to extend this result to a case where the matrix doesn't have integer coefficients but generating functions (with real coefficients) instead and to do so, I'd be interested in having a closer look at the proof of the standard case.

The questions I'm trying to figure out are:

1) is there a standard proof in a general case where we're dealing with a ring with a norm, than the geometric series has 1 as radius of convergence? Or what's the most obvious proof in the specific case of symmetric non-negative matrices?

2) what is the good equivalent of the case of matrices with integer coefficients that aren't necessarily symmetric. I'm quite interested in keeping the link to the eigenvalues and not in other matrix norms.

3) when dealing with matrices of generating functions (on one variable for a start), can I think of the eigenvalues as generating functions as well? Would that be the good way to think of this result on geometric series?

I'll keep posting as I work on all this (but I needed a place to write things down and possible get some help or hints).

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I don't know how exactly you want to make the generating functions into a normed algebra (it should be complete), but what you need for the usual argument to work is that $\lim_{n \to \infty} \Vert A^{n} \Vert^{1/n} \lt 1$, where you consider $A$ as an element of the continuous endomorphisms. The norm on the algebra of generating functions should satisfy $|a \ast b| \leq |a| \,|b|$ and the norm on the endomorphisms should be the usual sup-norm. I don't see why symmetry should be of any relevance. See also the Wikipedia page on the Neumann series. – t.b. May 24 '11 at 0:21
Your question is unclear. Matrices don't have coefficients; they have entries. Do you really want a matrix who entries are generating functions? – Gerry Myerson May 24 '11 at 1:05
Sorry, I did mean entries, not coefficients; so yes, I'm looking at matrices whose entries are GF. – David Kohler May 24 '11 at 14:58
up vote 3 down vote accepted

In any Banach algebra with identity over $\mathbb C$, the spectral radius of $x$ is $\rho(x) = \lim_{n \to \infty} \|x^n \|^{1/n}$. If $|t| \rho(x) < 1$, the geometric series $\sum_{n=0}^\infty t^n x^n$ converges to $(1 - t x)^{-1}$. This is standard material that you can find in any text that discusses Banach algebras.

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This is what I was looking for, thanks Robert. – David Kohler May 24 '11 at 14:56

As stated in my comment, the question is not clear, so this answer may not be relevant, but maybe it will be of some use.

Suppose $A$ is a square matrix with complex entries. Then there's an invertible matrix $P$ with $P^{-1}AP=J$ in Jordan form. Then $$\sum A^i=\sum(PJP^{-1})^i=P\left(\sum J^i\right)P^{-1}$$ so $\sum A^i$ converges if and only if $\sum J^i$ converges. But it's very easy to work out what $J^i$ looks like, first if $J$ is a single Jordan block, then for general $J$ in Jordan form, and thus work out conditions for convergence. The special case in which $A$ is diagonalizable is especially easy.

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