complex eigenvalue of self-adjoint matrix

Question

I have to diagonalize self-adjoint matrix above. But I cannot solve the characteristic polynomial, which is cubic equation with not noticeable factor. So when I insert that equation into wolfram alpha, it gives complex value. But thats contradictory, I think, because I have learned that all eigenvalue of Self-adjoint matrix are real. What is wrong here?

Answer 1

It simply means you made a careless mistake and you encounter a bug at that particular moment.

I inserted the matrix into wolfram alpha and the charactheristic polynomial is

$$-\lambda^3+7\lambda^2-9\lambda - 1$$

and I obtain $3$ distinct real eigenvalues.

Answer 2

The polynomial $-\lambda^3 + 7 \lambda^2 - 9 \lambda - 1$ does indeed have three real roots, approximately $-.1027750491$, $1.853634511$, $5.249140538$. Being a cubic, it can be solved in "closed form" using radicals. However, this is the "casus irreducibilis": these radicals must involve complex-valued intermediate expressions, even though the end result is real. If you try to evaluate those expressions numerically, there is a good chance that roundoff error will result in small, but nonzero, imaginary parts.

Alternatively, you can express the roots using trigonometric and inverse-trigonometric functions, and these are manifestly real:

$$ \frac{7}{3} + \frac{2 \sqrt{22}}{3} \cos \left(\frac{\arctan(3 \sqrt{237}/23)+ 2 k \pi}{3} \right),\ k = 0,1,2 $$