Discriminant of Characteristic Polynomial > 0

[jrcdude]

Homework Statement

Show that the descriminant of the characteristic polynomial of K is greater than 0.

K=\begin{pmatrix}-k_{01}-k_{21} &amp; k_{12}\\<br /> k_{21} &amp; -k_{12}<br /> \end{pmatrix}<br />

And k_i &gt; 0

Homework Equations

b^2-4ac&gt;0

The Attempt at a Solution

I have tried the following:
<br /> \begin{pmatrix}-k_{01}-k_{21}-\lambda &amp; k_{12}\\<br /> k_{21} &amp; -k_{12}-\lambda<br /> \end{pmatrix}<br />

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})&gt;0

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

Any help would be appreciated, thanks.

 

[Dick]

Homework Statement

Show that the descriminant of the characteristic polynomial of K is greater than 0.

K=\begin{pmatrix}-k_{01}-k_{21} &amp; k_{12}\\<br /> k_{21} &amp; -k_{12}<br /> \end{pmatrix}<br />

And k_i &gt; 0

Homework Equations

b^2-4ac&gt;0

The Attempt at a Solution

I have tried the following:
<br /> \begin{pmatrix}-k_{01}-k_{21}-\lambda &amp; k_{12}\\<br /> k_{21} &amp; -k_{12}-\lambda<br /> \end{pmatrix}<br />

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})&gt;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.

 

[jrcdude]

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.