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$

when i tried to compute it using gram schmidt, when $v_2=x_2-\langle x_2,x_1\rangle x_1$ here produce $0$? but all the vector seem already linear independent to each other?

  three of vector that i choose are 

$x_1=\begin{eqnarray*} \begin{bmatrix} \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \\ \end{bmatrix},x_2= \begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix} \text{ for } \lambda=-1,x_3= \begin{bmatrix} -1\\ 0 \\ 1 \\ \end{bmatrix}\text{ for } \lambda=-1, \end{eqnarray*}$

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 gram schmidt,$ \begin{eqnarray*}\frac{1}{\sqrt 3}\begin {bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix}. \begin {bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}\end{eqnarray*}$ it gave zero whats wrong?? and there seem many combination that i can choose as my eigen vector for $\lambda=-1$, what is the rule for that?

  • $\begingroup$ $x_1$ belongs to a different eigenspace than the other two vectors, so you already know that it’s going to be orthogonal to them. There’s no particular good reason to include it in the orthogonalization process. You just need to find an orthonormal pair that spans the eigenspace of $-1$. $\endgroup$ – amd Dec 3 '17 at 23:27
  • $\begingroup$ @amd to get orthonormal pair for $-1$ is it means i have to apply gram schmidt twice to $x_2$ and $x_3$ right? $\endgroup$ – fiksx Dec 4 '17 at 6:29
  • $\begingroup$ That’s right, although you really only need to normalize $x_2$. $\endgroup$ – amd Dec 4 '17 at 20:15
  • $\begingroup$ Also, since you’re working in $\mathbb R^3$, you could cheat and take the cross product of $x_1/\|x_1\|$ and $x_2/\|x_2\|$ for the third eigenvector. $\endgroup$ – amd Dec 4 '17 at 20:25

There is nothing wrong. It happens that your vectors $x_1$ and $x_2$ are already orthogonal. So, if you apply Gram-Schmidt to $\{x_1,x_2\}$, then, since $\|x_1\|=1$, what you get is $\left\{x_1,\frac{x_2}{\|x_2\|}\right\}$. Actually, what you have to do is to apply Gram-Schmidt to $\{x_1,x_2,x_3\}$ but, since $x_1$ is already orthogonal to the other two, just apply Gram-Shmidt to $\{x_2,x_3\}$ and then add $x_1$ to what you got.

  • $\begingroup$ thankyou, but here did you mean i should apply gram schmidt to $x_2 $ & $x_3$? $\endgroup$ – fiksx Dec 3 '17 at 15:44
  • $\begingroup$ @Vixf Indeed, that's what I meant. Since they are linearly independent, there«s no problem with that. $\endgroup$ – José Carlos Santos Dec 3 '17 at 15:45
  • $\begingroup$ thankyou so much!! after some calculation, i got $x_1=\begin{eqnarray*} \begin{bmatrix} \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \\ \end{bmatrix},x_2= \begin{bmatrix} 0 \\ \frac{-1}{\sqrt2} \\ \frac{1}{\sqrt2} \\ \end{bmatrix},x_3= \begin{bmatrix} \frac{-2}{\sqrt6}\\ \frac{1}{\sqrt6}\\ \frac{1}{\sqrt6} \\ \end{bmatrix}, \end{eqnarray*}$ but is this same as if i change $x_3$ to$x_3=\begin{eqnarray*} \begin{bmatrix} \frac{-1}{\sqrt6}\\ \frac{-1}{\sqrt6}\\ \frac{2}{\sqrt6} \\ \end{bmatrix}\end{eqnarray*}$ ? $\endgroup$ – fiksx Dec 3 '17 at 16:16
  • $\begingroup$ and also after i applied gram schmidt to $x_2$ & $x_3$, i need to applied it again to find the $x_3$ since it is still not orthogonal to the other vector right? $\endgroup$ – fiksx Dec 3 '17 at 16:20
  • $\begingroup$ @Vixf I don't understand any of your questions. The vectors$$x_1=\begin{bmatrix}\frac1{\sqrt3}\\\frac1{\sqrt3}\\\frac1{\sqrt3}\end{bmatrix}\,\ x_2=\begin{bmatrix}0\\\frac{-1}{\sqrt2}\\\frac1{\sqrt2}\end{bmatrix},\text{ and }x_3=\begin{bmatrix}\frac{-2}{\sqrt6}\\\frac1{\sqrt6}\\\frac1{\sqrt6}\end{bmatrix}$$are already an answer to your problem, although I don't understand how you got them (they're not what you would get applying Gram-Schmidt to the original vectors). Why would you want to change $x_3$? $\endgroup$ – José Carlos Santos Dec 3 '17 at 16:28

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