I have the matrix

$\displaystyle A=\left[ \begin{matrix} i & 1 \\ 0 & (-1+i) \end{matrix} \right]$

With the eigenvalues

$\displaystyle \lambda_1 = i, \lambda_2 = (-1+i)$

then the eigenvectors

$\displaystyle \lambda_1 : \left[ \begin{matrix} i - i & 1 \\ 0 & (-1+i) - i \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right]$

$\displaystyle \lambda_2 : \left[ \begin{matrix} i - (-1+i) & 1 \\ 0 & (-1+i) - (-1+i) \end{matrix}\right] \sim \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

Now I understand that for $\displaystyle \lambda_2$ the eigenvector is $\displaystyle \left[ 1 , -1\right]$ since $\displaystyle x_1 + x_2 = 0 \rightarrow x_1 = -x_2$.

However when it comes to the eigenvector for $\displaystyle \lambda_1$ all I have (in reduced form) $\displaystyle x_2 = 0$.

Why is the eigenvector for $\displaystyle \lambda_1 \rightarrow \left[ 1 , 0 \right]$?

Is it because the first column $\displaystyle (x_1)$ is a free variable?