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

I have some difficulty to solve this question:

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

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.

• Suba good answer. I used rlocusplot in Matlab and basically got the same answer. May 9 '18 at 16:28

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 $\alpha$.