I have a question from an exercise. I am given a vector space over the field $\mathbb{R}^{3}$ with 2 dimensions and I am asked to find a basis of eigenvectors. I found the eigenvalues but I have noticed whenever I tried to find the eigenvectors I would have a vector in $\mathbb{R}^{2}$, generally finding $\ker\left(xI-\left[T\right]_{B^{'}}\right)$ would give me the eigenvectors . So I didn't know how to solve it and this was the solution. Can someone explain me why the linear combinations are giving me here the eigenvectors at the end? Thank you!
Question:Let there be the vector space;
$V=\left\{ \left[\begin{array}{c} x\\ y\\ z \end{array}\right]\in\mathbb{R}^{3}|x+y+z=0\right\} $
and the linear transformation
$T\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=\left[\begin{array}{c} 3x+3y-3z\\ 4x-2y+4z\\ -x+5y+5z \end{array}\right]$
Find basis B of eigenvector and write $\left[T\right]_{B}$
Solution
We will chose arbitrary basis to calculate the eigenvalues.
$B^{'}=\left[\left[\begin{array}{c} 1\\ -1\\ 0 \end{array}\right],\left[\begin{array}{c} 1\\ 0\\ -1 \end{array}\right]\right]$
Then $\left[T\right]_{B^{'}}=\left[\begin{array}{cc} -6 & 0\\ 6 & 6 \end{array}\right]$
$\det\left|xI-\left[T\right]_{B^{'}}\right|=\left[\begin{array}{cc} x+6 & 0\\ -6 & x-6 \end{array}\right]=\left(x-6\right)\left(x+6\right)$
The eigenvalues are x=6 and x=-6
$x=6\Rightarrow \ker\left(6I-\left[T\right]_{B^{'}}\right)=\operatorname{span}\left(\left[\begin{array}{c} 0\\ 1 \end{array}\right]\right)$
$x=-6\Rightarrow \ker\left(-6I-\left[T\right]_{B^{'}}\right)=\operatorname{span}\left(\left[\begin{array}{c} -2\\ 1 \end{array}\right]\right)$
Then our basis $B=\left(u_{1},u_{2}\right)$ of eigenvector are
$\begin{cases} u_{1}=0v_{1}+v_{2}\\ u_{2}=-2v_{1}+v_{2} \end{cases}\Rightarrow\left[\begin{array}{cc} 6 & 0\\ 0 & -6 \end{array}\right]$