## Discriminant of Characteristic Polynomial > 0

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.

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 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.
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.

 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.
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