On Jun 12, 10:24 pm, Aaron Meurer <asmeu...@gmail.com> wrote:
> If the eigenvalues cannot be expressed in radicals, then it doesn't
> matter what method you use to compute them.
What about symbolic elements ? Or exact elements ?
>
> And for whatever reason, people always seem to be confused about this.
> The general fifth order and higher polynomial does not have a
> solution in radicals, and you can construct specific fifth order and
> higher polynomials whose roots are not expressible in radicals (like
> x**5 - x + 1). But that doesn't mean that *all* fifth order
> polynomials don't have solutions in radicals. It's very easy to
> construct a polynomial of any degree that has solutions in radicals.
> For example (x - 1)**n is a nth degree polynomial, and the roots are
> all 1. It's even possible to have an irreducible polynomial of degree
> 5 or greater whose solution is expressible in radicals.
You would find that you were wrong in your last statement. If a
polynomial has solutions in radicals, then it is necessarily a
reducible polynomial. If x0, x1, x2, ... are the exact solutions in
radicals then (x - x0)(x - x1)... is the reduced polynomial.
But we don't want to discuss the theoretical subtleties of the abel-
ruffini theorem.
My point is that we can't be sure that an arbitrary matrix will have
such a well-conditioned equation like (x - a)**n. Thus the charpoly
method (berwkowitz OR det(A-x*I), doesn't matter) is not a good
method, as it is not reliable for n > 5.
The method will not return with an exact eigenvalue for large
matrices.
>
> So there's no reason to just "give up" if the polynomial is degree 5
> or higher unless you are always solving the general equation. You can
> easily create a square matrix of any size whose eigenvalues are easily
> expressed (in radicals, or for example just integers). Yes, there
> will be cases where it can only produce roots in the form of RootOf,
> but either just let those pass through, or raise the error only when
> that happens (depending on if it works to just let them pass through).
RootOf is a good method, but only if the eigenval is the final result
desired by the user. What if the user wants the eigenvectors or the
SVD ?
Maybe we could follow the wolfram-alpha example. It returns with "root
of <charpoly> near <approximate solution>".
So, that RootOf objects could return the approximate value if
explicitly asked for.
>
> And by the way, maybe you were thinking that the characteristic
> polynomial isn't computed by det(A - x*I), as is often taught in
> linear algebra courses?
Didn't get you here. det(A - x*I) or berkowitz both give a charpoly.
So it wouldn't matter which we use, other than the fact that one is
faster than the other.
>
> Aaron Meurer
>
>
>
>
>
>
>
> On Sun, Jun 12, 2011 at 6:49 AM, SherjilOzair <sherjiloz...@gmail.com> wrote:
>
> > On Jun 12, 3:09 am, Vinzent Steinberg
> > <vinzent.steinb...@googlemail.com> wrote:
> >> On 11 Jun., 10:47, SherjilOzair <sherjiloz...@gmail.com> wrote:
>
> >> > Do you require to solve eigenvalue problems of matrices bigger than
> >> > 4*4 ?
> >> > How are you doing it currently ?
> >> > Matrix.diagonalize only works for matrices smaller than 5*5, as
> >> > polys.roots can only solve degree 4 equations and less.
>
> >> This is not entirely true, because we can find roots of higher order
> >> using for examples factorization. But yes, for equations of degree
> >> greater than 4 there is no general algorithm. But this does not matter
> >> in this case, because sympy does not calculate eigenvalues using the
> >> characteristic polynomial AFAIK.
>
> > Arbitrary higher-order equations can not be solved.
> > And characteristic polynomial of an arbitrary matrix is arbitrary.
> > This means, that we can't use that method to compute eigenvals
> > reliably.
> > I don't think there are any other direct methods, though.
>
> > sympy uses this method.
>
> > def berkowitz_eigenvals(self, **flags):
> > """Computes eigenvalues of a Matrix using Berkowitz method.
> > """
> > return roots(self.berkowitz_charpoly(Dummy('x')), **flags)
>
> > eigenvals = berkowitz_eigenvals
>
> > Or are you referring to something else ?
>
> >> Vinzent
>
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