# Solving a system of linear equations to find two eigenvectors.

So I have \begin{align*} x - 2y + z & = 0 \\ -2x + 4y - 2z & = 0 \\ x - 2y + z & = 0 \end{align*}

I know I need to find two eigenvectors for the eigenspace with eigenvalue 2 as I know the matrix is diagonalizable and I've already found the eigenvector for the eigenspace with eigenvalue 8...

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 Type this on WolframAlpha: Eigensystem[{{1,-1,1},{-2,4,-2},{1,-2,1}}], where are you getting stuck if you found the eignevalues and one of the eigenvectors? – Amzoti Dec 2 '12 at 23:36 I need the solutions for this system of equations as I can't work it out for some reason. – Jordan Dec 2 '12 at 23:43 y = 0, z = - x (clear?). Does that help? What type of matrix is this? Please describe what you have done? eigenvalue = 8? Did you see the eigenspace that WA produced (I gave you the command). I am trying to guide you because I don't want to spoon feed answers as that is not helpful to you. – Amzoti Dec 2 '12 at 23:54 Yeah, I guess you meant(1,-2,1) in the first part of the WA command? Yeah, I get it when y = 0, z = -x and that's a solution? And if z=0, x=2y and if x=0, z = 2y? – Jordan Dec 2 '12 at 23:58 Try watching this and see if it helps: khanacademy.org/math/linear-algebra/v/… – Amzoti Dec 2 '12 at 23:58
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\begin{align*} x - 2y + z & = 0 \\ -2x + 4y - 2z & = 0 \\ x - 2y + z & = 0 \end{align*}

\implies \begin{align*} x - 2y + z & = 0 \\ 0 & = 0 \\ 0 & = 0. \end{align*}

So you have two free variables. Assuming $y=t, z=s,\, t,s\in \mathbb{R}$, then the solution is given by

$$\begin{bmatrix} x \\ y\\ z\end{bmatrix} = \begin{bmatrix} 2t-s \\ t\\ s\end{bmatrix} = \begin{bmatrix} 2t \\ t\\ 0\end{bmatrix} + \begin{bmatrix} -s \\ 0\\ s\end{bmatrix} = t\begin{bmatrix} 2 \\ 1\\ 0\end{bmatrix} + s\begin{bmatrix} -1 \\ 0\\ 1\end{bmatrix}.$$

Can you see the two eigenvectors now?

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