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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

• Follow-Ups:
  • Re: Re: Help with Root function
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>