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The matrix associated with $f$ is: $$ \left(\begin{array}{rrr} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{array}\right) . $$

First, I am going to find the characteristic equation of $\det(A- \lambda I)$. Please correct me if I'm wrong.

$$= (3-\lambda)(3-\lambda)(3-\lambda)-1-1-(3-\lambda)-(3-\lambda)-(3-\lambda) =-\lambda^3+9\lambda^2-24\lambda+16 .$$

How to factor this? I know the $\lambda$'s should be $1$, $4$ and $4$. But how am I supposed to find these values?

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    One thing I always try is the [rational root theorem](http://en.wikipedia.org/wiki/Rational_root_theorem). This would identify $1$ as a root in short order, and then you can divide by $\lambda - 1$ to get a quadratic, which you can definitely factor. There might be more intelligent ways, though.2011-12-04
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    There is nothing wrong with guessing the roots of a polynomial. The best approach is plotting it.2011-12-04
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    Your title was incorrect. You don't want to compute the characteristic polynomial of $\det(A-\lambda I)$, you are trying to compute the characteristic polynomial of $A$. Also, you are not trying to find "the characteristic equation of $\det(A-\lambda I)$"; the characteristic equation of $A$ *is* $\det(A-\lambda I) = 0$.2011-12-04

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