
Eigenvectors
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?

Re: Eigenvectors
because any vector of the form (a,0) = a(1,0) satisfies x_{2} = 0, so this is an eigenvector iff a ≠ 0.
yes this is the same as being a "free variable", any choice of x_{1} (freely made) will do (but of course, the fact that an eigenvector must be nonzero means we can't choose x_{1} = 0, although (0,0) *is* in the eigenspace E_{λ1}).

Re: Eigenvectors
First, do you understand that there is no such thing as "the eigenvector" for given eigenvalue? Given any eigenvector, any (nonzero) multiple of it is also an eigenvector. In fact, the set of all eigenvectors for a given eigenvalue form a subspace. You say that "the eigenvector" for eigenvalue $\displaystyle \lambda_2$ is [1, 1]. In fact, any vector of the form [a, a], for a any (nonzero) number is an eigenvector.
I had to include "(nonzero)" because Deveno says "but of course, the fact that an eigenvector must be nonzero". Some text (but, admittedly a minority) define an eigenvalue by saying "$\displaystyle \lambda$ is an eigenvalue for linear operator A if and only if there exist a nonzero vector $\displaystyle Av= \lambda v$" but then define an "eigenvector" as any vector satisfying $\displaystyle Av= \lambda v$. That has the advantage, in my opinion, of including the 0 vector as an "eigenvector" so that the set of all eigenvector corresponding to a given eigenvalue is a subspace without having to add the 0 vector separately.