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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 8.2012-12-05

2 Answers 2