Discriminant of Characteristic Polynomial > 0

  • Thread starter jrcdude
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  • #1
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Homework Statement


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]

Homework Equations



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

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


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]

Homework Equations



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

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.

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:
  • #3
16
0
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.
 

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