limits on symbol eigenvalues?

[Uwe Brauer]

Hello

I just started using mathematica. When I tried to calculate the
symbolic eigenvalues of a 16x16 matrix mathematica told me it couldn't

Is there a restriction?

Thanks

Uwe Brauer

[Curt Fischer]

I think Mathematica couldn't do it because it is impossible to find the
symbolic roots to a sixteenth order polynomial, in general.

--
Curt Fischer

[Yasvir Tesiram]

On Jun 4, 2004, at 3:49 AM, Uwe Brauer wrote:

> Hello
>
> I just started using mathematica. When I tried to calculate the
> symbolic eigenvalues of a 16x16 matrix mathematica told me it couldn't
>
> Is there a restriction?
>

Yes, memory usually.

Yas

[Andrzej Kozlowski]

On 5 Jun 2004, at 20:18, Curt Fischer wrote:

> I think Mathematica couldn't do it because it is impossible to find the
> symbolic roots to a sixteenth order polynomial, in general.
>
> --
> Curt Fischer
>
>
>

Actually this is not quite true. It is certianly possible to find
symbolic roots of a polynomial od degree 16 or higher and Mathematica
can do it. (Just try it yourself). What is not possible, in general, is
to express this solution in terms of radicals, but that is quite a
different issue, unrelated to this problem.

The original question can't be answered with so little information.

Andrzej Kozlowski

[norooz...@gmail.com]

Hi,
I just start to use Mathematica, and I have exactly same problem
I really appreciate If can give me some advice.
thanks
Leyla

[David Bailey]

Not every symbolic problem that you can pose has a symbolic solution.
For example, some symbolic integrals don't have symbolic solutions -
likewise for differential equations.

A symbolic eigenvalue problem of order N involves solving an N'th order
polynomial equation. Specific cases can be solved, but the general case
cannot be solved for N>=5. This restriction can in theory be relaxed (I
am not sure by how much) by the use of theta functions, though the
symbolic answers are impossibly large.

Even when a symbolic solution is possible, it may not be desirable
because it is excessively complicated, and possibly numerically unstable
if the coefficients are subsequently replaced by numbers. To see what I
mean, try evaluating:

Solve[a x^4 + b x^3 + c x + d == 0, x]

David Bailey
http://www.dbaileyconsultancy.co.uk