# Question about different eigenvalues from the same matrix

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?

The two different answers you have found are equivalent $$(\frac{1 + \sqrt5}{2})^2 = \frac{6 +2\sqrt5}{4} = \frac{3 +\sqrt{5}}{2}$$

• Oh yeah, I didn' t see that - feel a little stupid right now :)
– John
Feb 27, 2017 at 13:44
• Its ok, it happens to most of us haha. Feb 27, 2017 at 15:00

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...