# 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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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
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

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