<s>[INST] Using the spectral method to solve $Ax=b$ I cannot find where I have made a computational error in applying the spectral method for solving $Ax=b$.
I am given $A = \begin{pmatrix} 7 &-2 & 1 \\ -2 & 10 & -2  \\ 1 & -2 & 7 \end{pmatrix}, b = \begin{pmatrix} 6 \\ 12 \\ 18 \end{pmatrix}$.
I've started by computing the eigenvalues and eigenvectors, I have the eigenvalues $r_1 = 12, r_2 = 6, r_3 = 6$ with  eigenvectors $v_1 = (1,-2,1), v_2 = (-1,0,1), v_3 = (2,1,0)$ respectively. 
Now, $x$, by the spectral method is given by 
$$\displaystyle x= \frac{u_1\cdot b}{r_1}u_1 + \frac{u_2\cdot b}{r_2}u_2+\frac{u_3\cdot b}{r_3}u_3.$$
Normalizing $v_1,v_2,v_3$ I get $u_1=\begin{pmatrix} \frac{1}{\sqrt{6}} \\ \sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{6}} \end{pmatrix} ,u_2 = \begin{pmatrix}\frac{-1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{pmatrix}, u_3 = \begin{pmatrix} \frac{2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \\ 0 \end{pmatrix}.$
So $x = 0 + \frac{6\sqrt{2}}{6}\left(-1/\sqrt{2},0,1/\sqrt{2}\right) + \frac{4}{\sqrt{5}}\left(2/\sqrt{5},1/\sqrt{5},0\right)=(7,4,1).$
This does not satisfy $Ax=b$ -- I can't find my error. By normal method of row reduction, I get $x=(1,2,3)$. Wolfram confirms this too! Am I incorrectly applying something here?
 [/INST] Your so-called spectral method works if the eigenvectors form an orthonormal basis, but this is not the case here: your $v_2,v_3$ are not even mutually orthogonal.
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