0
$\begingroup$

I have some difficulty to solve this question:enter image description here

What is the condition for real eigen values other than symmetric matrix?

$\endgroup$
2
$\begingroup$

A graphical method is to find the characteristic equation and then determine the threshold values using the root locus plot.

Here are the calculations using Mathematica. The largest value is 5.75.

enter image description here

$\endgroup$
1
  • $\begingroup$ Suba good answer. I used rlocusplot in Matlab and basically got the same answer. $\endgroup$ May 9 '18 at 16:28
2
$\begingroup$

At the bifurcation between all real eigenvalues and one pair of complex conjugate eigenvalues there has to be a duplicate real eigenvalue. So the characteristic polynomial of the matrix at the bifurcation should be of the following form

$$ (\lambda - a)^2(\lambda - b)(\lambda - c) = \lambda^4 - \left(2\,a + b + c\right)\,\lambda^3 + \left(2\,a\,(b + c) + b\,c + a^2\right)\,\lambda^2 - \left(a^2\,(b + C) + 2\,a\,b\,c\right)\,\lambda + a^2\,b\,c. $$

The actual characteristic polynomial of the matrix can be shown to be equal to

$$ \det(\lambda\,I - A) = \lambda^4 + 0\,\lambda^3 - 15\,\lambda^2 + 10\,\lambda + 24 + \alpha. $$

So in order for these two to be equal the following is required

$$ \begin{align} 2\,a + b + c &= 0 \\ 2\,a\,(b + c) + b\,c + a^2 &= -15 \\ a^2\,(b + c) + 2\,a\,b\,c &= -10 \\ a^2\,b\,c &= 24 + \alpha \end{align} $$

simplifying these equations gives

$$ \begin{align} b + c &= -2\,a \\ b\,c &= 3\,a^2 - 15 \\ 2\,a^3 - 15\,a + 5 &= 0 \\ 3\,a^2\,(a^2 - 5) &= 24 + \alpha \end{align} $$

In order for $b$ and $c$ to be real it can be shown that it is required that $2\,a^2 < 15$ and for $\alpha$ to be positive it is required that $2\,a^2 > 5 + \sqrt{57}$. By checking the roots of the cubic equation in $a$ against these constraints and using those can be used to solve for $\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.