NetNews Usenet Archive 1992 #18/NN_1992_18.iso/spool/sci/math/10201

Text File | 1992-08-12 | 2.8 KB | 58 lines

Newsgroups: sci.math
Path: sparky!uunet!usc!sol.ctr.columbia.edu!destroyer!ubc-cs!unixg.ubc.ca!unixg.ubc.ca!israel
From: [e-mail] (Robert B. Israel)
Subject: Re: Help - non-integral powers of a matrix?
Message-ID: <[e-mail]>
Keywords: matrix
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Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Thu, 13 Aug 1992 06:20:59 GMT
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In article <[e-mail]> [e-mail]
(John C. Baez) writes:

>In article <a_rubin.713653963@dn66> [e-mail] (Arthur
>Rubin) writes:
(referring to the series log A = sum_{k=1}^infinity (-1)^(k-1) (A-I)^k/k)
>>log "obviously" converges if all eigenvalues of A are strictly within 1 of 1,

>To clarify, perhaps, let me add that this is not only "obvious", it's
>true, at least if A is diagonalizable.  If A is a
>not-necessarily-diagonalizable matrix log A is defined if ||A - 1|| < 1;
>I don't think the eigenvalue condition above is sufficient.

It is sufficient.  On the other hand, |z - 1| <= 1 with z <> 1 (which I put
forward in another post) is NOT enough.

We can assume without loss of generality that A has only one eigenvalue r.
Then A = rI + N where N is nilpotent (say N^n = 0).  Expanding all the terms
of our series in powers of N, it is enough to get convergence for the
coefficients of each N^j, 0 <= j < n.  The series for the coefficient of N^j
is
    sum_{k = j}^infinity (-1)^(k-1) binom(k,j)/k (r-1)^(k-j)
= ((r-1)^(-j)/j!) sum_{k = j}^infinity (-1)^(k-1) (k-1)(k-2)...(k-j+1) (r-1)^k
and this converges if |r - 1| < 1.

Let's make a clear distinction between "log A is defined" (i.e. there is
some (highly non-unique) B such that exp(B) = A) and "the above series
converges".  For log A to exist, all you need is that A is nonsingular.
Moreover, you can get log A without any infinite series.

If, as above, A = rI + N with N^n = 0 and r <> 0, then
    log A = log r + log(I + N/r)
          = log r + sum_{k=1}^{n-1} (-1)^(k-1) (N/r)^k /k
In general, for any nonsingular A, decompose the space of column vectors
into eigenspaces corresponding to each eigenvalue; on each eigenspace, A has
the above form.  Of course, finding the eigenvalues is difficult and
numerically unstable in general, so this is more of a theoretical than a
practical solution.


--
Robert Israel                            [e-mail]
Department of Mathematics             or [e-mail]
University of British Columbia
Vancouver, BC, Canada V6T 1Y4