One eigenvalue is 2.
A= |-1 3| |2 0|
-3x+3y=0
2x-2y=0
x=y
I don't know the answer
In fact, the set of all vectors v such that $\displaystyle Ax= \lambda x$, for $\displaystyle \lambda$ and eigenvalue of a A is a subspace. That's why you cannot solve for just a single pair of values for x and y. You arrived at y= x so <x, y>= <x, x>= x<1, 1> or all mutiples of <1, 1> are eigenvectors corresponding to eigenvalue 2. That is the "eigenspace" for eigenvalue 2.