Let $A=\begin{bmatrix}2&1&2\\4&2&4\\2&1&2\end{bmatrix}$
Its eigenvalues are $0,0,6$. I want to find its eigenvectors.
My solution:
when $\lambda_1=0$
$(A-\lambda_1I)x_1=\begin{bmatrix}2&1&2\\4&2&4\\2&1&2\end{bmatrix}x_1=0$
$x_1=(-\frac{1}{2}s-t s t)=-\frac{1}{2}s(1, -2, 0)+t(-1, 0, 1)$ since 2nd and 3rd vairables are free variables. Thus $(1, -2, 0)$ and $(-1, 0, 1)$ are the eigenvectors.
BUT the answer is $(1, -2, 0)$ and $(0, -2, 1)$.
I can't understand why these are the true answers. What is wrong with my solution?