From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: necessary condition for series convergence
Date: 21 Dec 1994 06:22:59 GMT

In article <3d79hn$s7t@newstand.syr.edu>, hgraber@lynx.cat.syr.edu (Harry
Graber) asked about the convergence of a matrix power series
> 
>                 Inv(I - A) = I + A + A^2 + A^3 + ...

In article <edgar-2012941445160001@dizzy.mps.ohio-state.edu>,
Gerald Edgar <edgar@math.ohio-state.edu> wrote:
>It converges if all eigenvalues of A are <1 in absolute value.
>It diverges if some eigenvalue is >1 is absolute value.
>If some eigenvalues are exactly =1 in absolute value, it may or
>may not converge.
 
Are you using a generous notion of convergence here? I'd say it really
diverges unless every eigenvalue is less than 1 in magnitude.

Really I wouldn't bother following up except to say a few words that might
be helpful before anyone else tries matrix series. The trick to
understanding these issues is to notice that conjugacy (by a single, fixed,
invertible matrix) preserves sums, powers, scalar multiples, and limits.
For example we have  Inv(I-B^(-1)AB) = B^(-1) Inv(I-A) B and
I + B^(-1)AB + (B^(-1)AB)^2 + ... = B^(-1) ( I + A + A^2 + ... ) B ,
so your equation holds for  A  iff it holds for  B^(-1)AB. This is good to
know because given any  A  we can choose  B  so that  B^(-1)AB  has,
say, the Jordan canonical form.

Now, if  f  is a power series and  A  is an upper-triangular matrix with
diagonal entries  a_1, ..., a_n,  then it's easy to see that the  i-th
term on the diagonal is just  f(a_i); more precisely, if the power series
f(A)  is to converge (in the sense that for each  i  and  j  the (i,j)
entires of the partial sums converge as a sequence of complex numbers) then
(taking j=i) the power series  f(a_i)  must all converge too. The converse
is trickier, although most matrices are diagonalizable and for diagonal
matrices the situation does reduce completely to the convergence on the
diagonal. This reduces 'radius of convergence' questions to the 
corresponding question for an ordinary power series.

For the power series  f(x) = 1 + x + x^2 + ... we get divergence for
x=1 because the partial sums are unbounded, and also divergence for other
numbers  x  of magnitude one because the partial sums  (1-x^n)/(1-x)
keep wrapping around a circle.

dave