How to find the square root
TCL
matrix.
Is there a quick way to find the square root of A without first
diagonalizing E?
-TCL
ksoileau
Using the binomial expansion for exponent 1/2, it is possible to write
the square root of I+E^2 as a series of powers of E^2. For example,
(1+1/2*E^2-1/8*E^4+1/16*E^6-5/128*E^8+7/256*E^10)^2-1-E^2=21/512*E^12
+ higher powers of E^2.
ksoileau
Correcting a few typos yields:
Using the binomial expansion for exponent 1/2, it is possible to
write
the square root of I+E^2 as a series of powers of E^2. For example,
(I+1/2*E^2-1/8*E^4+1/16*E^6-5/128*E^8+7/256*E^10)^2-I-E^2=21/512*E^12
Robert Israel
See e.g. the sci.math.num-analysis thread "Square root of a matrix?" from
September 2000
<http://groups.google.ca/group/sci.math.num-analysis/browse_thread/thread/f23513b68930e228>
and references there.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Herman Rubin
In article <4b693082-994f-41a0...@f6g2000vbp.googlegroups.com>,
There are many ways, but I am not sure of quick ones.
One could start with an initial value, such as (I+E)/sqrt(2),
whose characteristic roots with respect to the desired value
are between 1/sqrt(2) and 1, and use Newtons method, which
converges fairly rapidly. Starting with I+E is about as good,
and has the advantage that the approximants decrease.
Or one can start with I+E and use power series.
If one has a polynomial equation satisfied by E
of low degree, that can be used without diagonalization
to get a polynomial for sqrt(I+E^2) (or any other
function of E) aa polynomial of the same degree.
This method clearly also works if the characteristic
roots are known, multiplicity ignored.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558