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I am given the following matrix $$\left( \begin{matrix} 0 &1&0\\0&0&1\\1&-1&0 \end{matrix} \right).$$ Now I computed the characteristic polynomial of $A$ to be $p_A(\lambda) = \lambda^3+\lambda-1.$ This means that all eigenvalues are positive and thus the system $\dot{x} = Ax+Bu$ is unstable. Now I wonder if there is any way someone can find the exact eigenvalues, I tried to factorize the characteristic polynomial endlessly but couldn't get it. Thanks in advance.

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  • $\begingroup$ @Ayman Hourieh, thanks for the edit. $\endgroup$ – Slugger Jan 23 '14 at 16:15
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The eigenvalues are the roots of the characteristic polynomial. A cubic can be solved using radicals, but the result is rather messy: $$ \frac16\,\sqrt [3]{108+12\,\sqrt {93}}-2\,{\frac {1}{\sqrt [3]{108+12\, \sqrt {93}}}}$$ $$-\frac{1}{12}\,\sqrt [3]{108+12\,\sqrt {93}}+{\frac {1}{\sqrt [3 ]{108+12\,\sqrt {93}}}}+\frac{i\sqrt {3}}{2} \left( \frac16\,\sqrt [3]{108+12\, \sqrt {93}}+2\,{\frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}} \right) $$ $$-\frac{1}{ 12}\,\sqrt [3]{108+12\,\sqrt {93}}+{\frac {1}{\sqrt [3]{108+12\,\sqrt { 93}}}}-\frac{i \sqrt{3}}{2} \left( \frac16\,\sqrt [3]{108+12\,\sqrt {93}}+2\,{ \frac {1}{\sqrt [3]{108+12\,\sqrt {93}}}} \right) $$

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If you check again, you should find that the characteristic polynomial of $A$ is in fact $\lambda^3+\lambda+1.$ It isn't nicely factorable (though Cardano's method may be used to do so), but has one negative root and two complex conjugate imaginary roots.

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  • $\begingroup$ ah I cant believe I got that wrong, thanks for spotting that! $\endgroup$ – Slugger Jan 23 '14 at 16:51
  • $\begingroup$ You're welcome! $\endgroup$ – Cameron Buie Jan 23 '14 at 16:52

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