1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discriminant of Characteristic Polynomial > 0

  1. Nov 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the descriminant of the characteristic polynomial of K is greater than 0.

    [tex]K=\begin{pmatrix}-k_{01}-k_{21} & k_{12}\\
    k_{21} & -k_{12}
    \end{pmatrix}
    [/tex]

    And [itex]k_i > 0[/itex]

    2. Relevant equations

    [tex]b^2-4ac>0[/tex]

    3. The attempt at a solution

    I have tried the following:
    [tex]
    \begin{pmatrix}-k_{01}-k_{21}-\lambda & k_{12}\\
    k_{21} & -k_{12}-\lambda
    \end{pmatrix}
    [/tex]

    Bringing me to
    [tex]\lambda^{2}+(k_{12}+k_{01}+k_{21})\lambda+k_{01}k_{12}=0[/tex]

    And then plugging it into discriminant form

    [tex](k_{12}+k_{01}+k_{21})^{2}-4(k_{01}k_{12})>0[/tex]

    But from there I don't think that is a true statement.

    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Nov 26, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Actually, I think it is true. But it's not obvious. Let's call k12=x, k01=y and k21=z, so you want to show (x+y+z)^2-4yz>0 if x>0, y>0 and z>0. Just so we don't have to write the subscripts. I showed it by completing as many squares as I could in that expression after expanding it. Then it's easy to see.
     
    Last edited: Nov 26, 2012
  4. Nov 26, 2012 #3
    D'oh I think the form I was looking for was:

    [tex]x^2+2x(y+z)+(y-z)^2[/tex]

    which is clearly greater than zero.

    Thanks for the insight.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Discriminant of Characteristic Polynomial > 0
Loading...