Discriminant of Characteristic Polynomial > 0by jrcdude Tags: characteristic, discriminant, polynomial 

#1
Nov2612, 07:36 PM

P: 16

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^24ac>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
Nov2612, 08:48 PM

Sci Advisor
HW Helper
Thanks
P: 25,175





#3
Nov2612, 08:59 PM

P: 16

D'oh I think the form I was looking for was:
[tex]x^2+2x(y+z)+(yz)^2[/tex] which is clearly greater than zero. Thanks for the insight. 


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