1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

\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.

$\endgroup$
7
  • $\begingroup$ thanks but can you explain how to get this? is this using gram-schmidt? $\endgroup$
    – fiksx
    Dec 3 '17 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
    Dec 3 '17 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). $\endgroup$ Dec 3 '17 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
    Dec 3 '17 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
    Dec 3 '17 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.