Finding the eigenvector of a matrix $2\times 2$ in two ways? I need some help.
I dont understand why the eigenvector $v_1$ of the matrix \begin{pmatrix} 0 & 1 \\ -1 & -2 \end{pmatrix} is the vector $(1,-1)$ and no the vector $(-1,1)$.
$\lambda_1 = \lambda_2 = −1$

Note that the vector $(1,-1)$ is just $(-1,1)$ multiplied by $-1$. Indeed, since an eigenvector is a basis vector in the null space of $A-\lambda I$, any scalar multiple of an eigenvector is also an eigenvector.
Or in other words,
$$(A-\lambda I) v = 0 \implies \alpha (A-\lambda I)v  = 0 \implies (A-\lambda I)(\alpha v) = 0$$
for any non-zero scalar $\alpha$.
