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$B=\begin{bmatrix}3&-2&1\\\\-2&6&-2\\\\1&-2&3\end{bmatrix}$

so far I have found

$char(f)=\begin{bmatrix}3-x&2&1\\\\-2&6-x&-2\\\\1&-2&3-x\end{bmatrix}$

from this the equation is $-x^3+12x^2-44x+48$ which factorizes to $-(x-6)(x-4)(x-2)$ so $\lambda=2,4,6$

I'm not sure what to do next, do I try and find a basis $E_\lambda$

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You must have made a mistake, $B$ has an eigenvalue of multiplicity 2 at 2 and another at 8. –  copper.hat Dec 5 '12 at 0:11
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You should really do something about that low accept rate percentage of yours. –  DonAntonio Dec 5 '12 at 0:11
    
@DonAntonio What do you mean accept rate? –  Adam Dec 5 '12 at 0:25
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@Adam Your accept rate is the percentage of your questions where you accepted an answer by choosing a checkmark next to it. –  Phira Dec 5 '12 at 0:27
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2 Answers

You made mistake somewhere in finding eigenvalues, they should be 2 2 8...

After that, find your eigenvectors $v_1, v_2, v_3$ (corresponding to eigenvalues). Then there exists an invertible matrix $P = [ v_1\ v_2\ v_3 ] $ and a diagonal matrix $D$ with eigenvalues in the diagonal such that $P^{-1}BP =D$

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Solving $(B-8I)x=0$ is fairly straightforward, $x=(1, -2, 1)^T$ is one solution.

Solving $(B-2I)x = 0$ is also fairly straightforward, $x=(1,1,1)^T$ is fairly easy to spot, and is $x=(1,0,-1)^T$.

Try a basis of eigenvectors.

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