# Discriminant of Characteristic Polynomial > 0

by jrcdude
Tags: characteristic, discriminant, polynomial
 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. $$K=\begin{pmatrix}-k_{01}-k_{21} & k_{12}\\ k_{21} & -k_{12} \end{pmatrix}$$ And $k_i > 0$ 2. Relevant equations $$b^2-4ac>0$$ 3. The attempt at a solution I have tried the following: $$\begin{pmatrix}-k_{01}-k_{21}-\lambda & k_{12}\\ k_{21} & -k_{12}-\lambda \end{pmatrix}$$ Bringing me to $$\lambda^{2}+(k_{12}+k_{01}+k_{21})\lambda+k_{01}k_{12}=0$$ And then plugging it into discriminant form $$(k_{12}+k_{01}+k_{21})^{2}-4(k_{01}k_{12})>0$$ But from there I don't think that is a true statement. Any help would be appreciated, thanks.
 Quote by jrcdude 1. The problem statement, all variables and given/known data Show that the descriminant of the characteristic polynomial of K is greater than 0. $$K=\begin{pmatrix}-k_{01}-k_{21} & k_{12}\\ k_{21} & -k_{12} \end{pmatrix}$$ And $k_i > 0$ 2. Relevant equations $$b^2-4ac>0$$ 3. The attempt at a solution I have tried the following: $$\begin{pmatrix}-k_{01}-k_{21}-\lambda & k_{12}\\ k_{21} & -k_{12}-\lambda \end{pmatrix}$$ Bringing me to $$\lambda^{2}+(k_{12}+k_{01}+k_{21})\lambda+k_{01}k_{12}=0$$ And then plugging it into discriminant form $$(k_{12}+k_{01}+k_{21})^{2}-4(k_{01}k_{12})>0$$ But from there I don't think that is a true statement. Any help would be appreciated, thanks.
 P: 16 D'oh I think the form I was looking for was: $$x^2+2x(y+z)+(y-z)^2$$ which is clearly greater than zero. Thanks for the insight.