🚀 go-pugleaf

jonmach wrote:
...
> e.g. [ 2 1 ]
>      [ 3 4 ]
> 
> gives the polynomial x^2 - 6x + 5 = 0
> giving roots (x - 1) (x - 5)
> 
> As you state,
> 
> > The eigenvectors can then be written as (b, x-a) or (x-d, c).
> 
> Your solution provides me with 4 potential eigenvectors.
> 
> (1, 1-2) = (1, -1) IS an eigenvector for the eigenvalue 1.
> (1, 5-2) = (1, -3) IS NOT an eigenvector for the eigenvalue 5.

you made an error in the calculation

> (1-4, 3) = (-3, 3) IS NOT an eigenvector for the eigenvalue 1.

it IS.

> (5-4, 3) = (1, 3) IS an eigenvector for the eigenvalue 5.
> 
> "Alan Miller" <amiller@bigpond.net.au> wrote in message news:<iB8G7.310759$bY5.1235789@news-server.bigpond.net.au>...
> > jonmach wrote in message ...
> > >Hi All,
> > >
> > >I've trawled the net and have found some very comprehensive libraries
> > >and source for calculating the eigenvalues/eigenvectors of an nxn
> > >matrix.  However, this is very much overkill for what I need.
> > >
> > >I have a single 2X2 matrix holding real values and simply need to
> > >calculate the 2 eigenvalues and corresponding eigenvectors.  Is there
> > >a very simple algorithm for this that already exists in
> > >C/Fortran/Java/C++?
> > >
> > >Many many thanks in advance for any assistance on this.
> > >
> > >Jon
> >
> > The matrix:-
> > ( a  b )
> > ( c  d )
> > has eigenvalues obtained by solving (a-x)(d-x) - bc = 0 for x.
> > This is simply a quadratic equation, and the solutions are:
> > 2x = -(a+d) +/- sqrt[ (a-d)^2 + 4bc ]
> > The roots can of course be complex, if b & c have opposite signs
> > and are moderately large compared with the diagonal elements a & d.
> > The eigenvectors can then be written as (b, x-a) or (x-d, c).
> ...
-- 
G"unter Rote, Freie Universit"at Berlin, Institut f"ur Informatik