Re: Help with Root function
- To: mathgroup at smc.vnet.net
- Subject: [mg79483] Re: [mg79407] Help with Root function
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Fri, 27 Jul 2007 06:04:12 -0400 (EDT)
- References: <14196226.1185458345477.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
The Root object will do anything for you that a form in terms of radicals
can do... except LOOK like a radical form. That being said, you can
SOMETIMES get the radical form as follows:
ToRadicals@Eigenvalues@{{a, 1, 1}, {1, b, 1}, {1, 1, c}}
{1/3 (a + b + c) - (2^(
1/3) (-9 - a^2 + a b - b^2 + a c + b c - c^2))/(3 (54 + 2 a^3 -
3 a^2 b - 3 a b^2 + 2 b^3 - 3 a^2 c + 12 a b c - 3 b^2 c -
3 a c^2 - 3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3)) + (1/(
3 2^(1/3)))((54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 - 3 a^2 c +
12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3)),
1/3 (a + b +
c) + ((1 + \[ImaginaryI] Sqrt[3]) (-9 - a^2 + a b - b^2 + a c +
b c - c^2))/(3 2^(
2/3) (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 - 3 a^2 c +
12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3)) - (1/(
6 2^(1/3)))(1 - \[ImaginaryI] Sqrt[3]) (54 + 2 a^3 - 3 a^2 b -
3 a b^2 + 2 b^3 - 3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 -
3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3),
1/3 (a + b +
c) + ((1 - \[ImaginaryI] Sqrt[3]) (-9 - a^2 + a b - b^2 + a c +
b c - c^2))/(3 2^(
2/3) (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 - 3 a^2 c +
12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3)) - (1/(
6 2^(1/3)))(1 + \[ImaginaryI] Sqrt[3]) (54 + 2 a^3 - 3 a^2 b -
3 a b^2 + 2 b^3 - 3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 -
3 b c^2 +
2 c^3 + \[Sqrt](4 (-9 - a^2 + a b - b^2 + a c + b c -
c^2)^3 + (54 + 2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 -
3 a^2 c + 12 a b c - 3 b^2 c - 3 a c^2 - 3 b c^2 +
2 c^3)^2))^(1/3)}
Bobby
On Thu, 26 Jul 2007 04:28:38 -0500, jeremito <jeremit0 at gmail.com> wrote:
> I am trying to find the eigenvalues of a 3x3 matrix with non-numeric
> elements. This requires finding the roots of cubic polynomials.
> Mathematica can do this, but I know how to interpret its output. For
> example
>
> In[1]:= B = {{a, 1, 1}, {1, b, 1}, {1, 1, c}}
>
> Out[1]= {{a, 1, 1}, {1, b, 1}, {1, 1, c}}
>
> In[2]:= Eigenvalues[B]
>
> Out[2]= {Root[-2 + a + b + c -
> a b c + (-3 + a b + a c + b c) #1 + (-a - b - c) #1^2 + #1^3 &,
> 1], Root[-2 + a + b + c -
> a b c + (-3 + a b + a c + b c) #1 + (-a - b - c) #1^2 + #1^3 &,
> 2], Root[-2 + a + b + c -
> a b c + (-3 + a b + a c + b c) #1 + (-a - b - c) #1^2 + #1^3 &,
> 3]}
>
>
> How can I get Mathematica to give me the full answer? I know it is
> long and ugly, but at least I can do something with it. I can't do
> anything with what it gives me now. Does that make sense?
> Thanks,
> Jeremy
>
>
>
--
DrMajorBob at bigfoot.com
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