I am trying to understand how to find the eigenvectors of a matrix, but I can't quite grasp the last step of the process. I've looked through two different textbooks, and tried watching some of the Khan Academy videos as well as looking over some other posts here. None of it has made it any clearer for me.
To keep it simple, let's take this $2\times 2$ matrix (my problem is the same regardless of the size of the matrix):
$A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$
It has eigenvalues $\lambda_1=-2$ and $\lambda_2=3$, which I can easily find. For this example I will just focus on the first of these.
If I calculate $A-\lambda_1 I_2$ and reduce it to row-eschelon form I get
$A+2I_2 = \begin{bmatrix} 3 & 2 \\ 3 & 2 \end{bmatrix} \to \begin{bmatrix} 3 & 2 \\ 0 & 0 \end{bmatrix}$
The system
$\begin{bmatrix} 3 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Has solutions $3x=-2y$. According to Wolfram|Alpha, the eigenvector for $\lambda_1=-2$ is $(-2,3)$. These numbers clearly come from the solution I just found, that's plain to see, what I can't see is how and why I go from $3x=-2y$ to $(-2,3)$.
I would think the eigenvector should be $(\frac{-2}{3}, \frac{-3}{2})$ instead, why isn't it?
Also, Wolfram|Alpha only reports that one vector, but if I understand things correctly, the matrix has an infinite number of eigenvectors (the eigenspace), and the one Wolfram|Alpha reports is only the basis of this space. Is this correctly understood? Why does Wolfram|Alpha only report that one?