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symmetry matrix $\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right) $ one of eigen value is $\lambda_1=2$ and one of eigen vector is $x_1=\left(\begin{array}{ccc} \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \end{array}\right) $ then i found the two other eigen value is $\lambda_2=-1 $ & $\lambda_3=-1$ but the question want eigen vector that is orthogonal to each other and has magnitude $ |x_2|=|x_3|=1$
here how can i find the two other eigen vector that orthogonal to each other? using usual computation $x=-y-z, i cant find the vector that is orthogonal to each other and with magnitude 1. using trial and error?

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\begin{eqnarray*} \begin{bmatrix} \frac{2}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\ \end{bmatrix}, \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \\ \end{bmatrix} \end{eqnarray*}

Edit: Note that \begin{eqnarray*} \begin{bmatrix} 2 \\ -1 \\ -1 \\ \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ -1 \\ \end{bmatrix}, \begin{bmatrix} -1 \\ -1 \\ 2 \\ \end{bmatrix} \end{eqnarray*} are eigenvectors with eigenvalue $-1$ and between them they span a space of dimension $2$, so we need to find a linear combination that is orthogonal the first one \begin{eqnarray*} \begin{bmatrix} 2 \\ -1 \\ -1 \\ \end{bmatrix} \cdot ( \alpha \begin{bmatrix} 2 \\ -1 \\ -1 \\ \end{bmatrix}+ \beta \begin{bmatrix} -1 \\ 2 \\ -1 \\ \end{bmatrix}+ \gamma \begin{bmatrix} -1 \\ -1 \\ 2 \\ \end{bmatrix} ) . \end{eqnarray*} This gives $ 6 \alpha -3 \beta -3 \gamma=0$ and $ \alpha=1, \beta=-2, \gamma=0$ will do, and now you just need to normalise.

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  • $\begingroup$ thanks but can you explain how to get this? is this using gram-schmidt? $\endgroup$
    – fiksx
    Commented Dec 3, 2017 at 12:35
  • $\begingroup$ thanks but how did you get eigen vector? i got $\begin{eqnarray*} \begin{bmatrix} -1\\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}\end{eqnarray*}$ ? and why has to be ortogonal to first one? i have no clue :/ $\endgroup$
    – fiksx
    Commented Dec 3, 2017 at 13:28
  • $\begingroup$ The above arguement works just as well with the vectors \begin{eqnarray*} \begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ 0 \\ \end{bmatrix} \begin{bmatrix} -1\\ 0 \\ 1 \\ \end{bmatrix}, \end{eqnarray*}. The question requests an orthonormal basis, so we need to find two linear combinations of these vectors whose dot product gives zero (orthogonal). $\endgroup$ Commented Dec 3, 2017 at 13:43
  • $\begingroup$ when i tried to compute it using gram schmidt, when $v_2=x_2-<x2.x1>x1$ here $<x2.x1>$ produce $0$? and all the vector seem already linear independent to each other? $\endgroup$
    – fiksx
    Commented Dec 3, 2017 at 14:08
  • $\begingroup$ i choose $\begin{eqnarray*} x_1=\frac{1}{\sqrt 3}\begin {bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix}, x_2= \begin {bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}\end{eqnarray*}$ and using grram schmidt, $\begin{eqnarray*} \begin{bmatrix} -1\\ 1 \\ 0 \\ \end{bmatrix}.\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \end{eqnarray*}$ it gives zero $\endgroup$
    – fiksx
    Commented Dec 3, 2017 at 14:21

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