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enter image description here 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?

Septacle
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2 Answers2

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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:

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

Robert Israel
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  • One must admit that computing these values by hand would not be trivial... – Math1000 Oct 30 '17 at 05:16
  • Thanks. I didnt know about the casus irreducibilis. By the way, can u teach me how to get the final values? – Septacle Oct 30 '17 at 06:05
  • The sum of the three roots is $7$. $x = \lambda - 7/3$ satisfies $x^3 - \frac{22}{3} x + \frac{92}{27} = 0$. If $x = b \cos(t)$, $\cos(3t) = 4 \cos(t)^3 - 3 \cos(t) = (4/b^3) (x^3 - (3 b^2/4) x)$. So take $b = 2 \sqrt{22}/3$, and you get $\cos(3t) = (23/242) \sqrt{22}$. Thus $t = \pm (1/3) \arccos((23/242) \sqrt{22}) + 2 \pi k/3$. – Robert Israel Oct 30 '17 at 19:58
  • For some reason, Maple expressed this using $\arctan$ instead of $\arccos$. – Robert Israel Oct 30 '17 at 19:59
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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.

Siong Thye Goh
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  • Mathematica gives me hideous expressions for those values though, like this: $$-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{23+3 i \sqrt{237}}}{3\ 2^{2/3}}+\frac{7}{3}-\frac{11 \left(1-i \sqrt{3}\right)}{3 \sqrt[3]{2 \left(23+3 i \sqrt{237}\right)}}.$$ – Math1000 Oct 30 '17 at 04:49
  • Have you tried wolfram alpha? It should be a real number. I wonder if the expression can be simplifed to a real number. – Siong Thye Goh Oct 30 '17 at 04:57