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I recenctly came onto this situation where I needed to compute that charasteristic polynomial of this table:

$M=\begin{pmatrix}0&1&0&0\\ \:1&0&1&0\\ \:0&1&0&1\\ \:0&0&1&0\end{pmatrix}$.

Now, we have: $x(\lambda)=det(M-\lambda I)=...=(\lambda^2-1)^2-\lambda^2$. Now, we can have either one of the two:

  • $x(\lambda)=\lambda^4-3\lambda^2+1$, where I substitute $u=\lambda^2$, get a quadratic equation and find in the end the eigenvalues as: $\lambda_i=\pm \sqrt{\frac{3\pm \sqrt5}{2}}$
  • Or, and that's what drives me mad, use this: $a^2-b^2=(a+b)(a-b)$ and get: $x(\lambda)=(\lambda^2+\lambda-1)(\lambda^2-\lambda-1)$ and find the eigenvalues as: $\lambda_i=\frac{\pm1\pm \sqrt5}{2}$, which I believe are different from the ones above! My question is if this is possible or I don't know - have I done some mistake somewhere?
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The two different answers you have found are equivalent $$(\frac{1 + \sqrt5}{2})^2 = \frac{6 +2\sqrt5}{4} = \frac{3 +\sqrt{5}}{2}$$

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  • $\begingroup$ Oh yeah, I didn' t see that - feel a little stupid right now :) $\endgroup$ – John Feb 27 '17 at 13:44
  • $\begingroup$ Its ok, it happens to most of us haha. $\endgroup$ – Ziad Fakhoury Feb 27 '17 at 15:00
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Observe that

$$\pm\sqrt\frac{3\pm\sqrt5}2=\frac{\pm1\pm\sqrt5}2\iff\frac{3\pm\sqrt5}2=\frac{6\pm2\sqrt5}4$$

Now just make sure you get the different signs correctly...

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