I have found the eigen vaues, I also know that you can find the eigenvectors through a Gausian Jordan. -- x1, gauss jordan gives me rows(1 -1/3 ,, 0 0 ), so [a, b] = [1,3] For vector x2, GJ gives (1 -2/5 ,, 0 0 ), I would assume [a,b] = [2,5], but why did they choose to go with [-2,-5]. I don't get it?
A bigger picture is on this webpage if needed;
http://oi59.tinypic.com/2v7unw1.jpg
On
Remember that an eigenvector defines a subspace of the domain of the linear transformation. That subspace has many bases --- in fact, each nonzero multiple of the eigenvector is also a basis for the eigenspace. So both $[2,5]$ and $[-2,-5]$ are bases, and either can be thought of as "representing" the one-dimensional eigenspace. The choice of $[-2,-5]$ is just that --- an arbitrary choice. Another valid choice would have been $[4,10]$ (twice $[2,5]$), or $[1,\frac{5}{2}]$.
Note that if the eigenspace has dimension greater than one, then this question appears to look slightly different. Suppose for example that $[1,0,0]$ and $[0,1,0]$ were eigenvectors. Then the eigenspace they define is the $xy$-plane. So an equally valid pair of eigenvectors defining that plane would be any other pair of vectors forming a basis. For example, $[-1,2,0]$ and $[3,5,0]$.
Any multiple of an eigenvector is still an eigenvector for the same eigenvalue, even if this multiple is negative. So if $(2,5)^T$ is an eigenvector then so are $(-2,-5)^T$, $(10,25)^T$, $(1,5/2)^T$, $(-6,-15)^T$, ...
If the machine marking the answers is clever enough (for example it's using MapleTA) it should accept any of these.
As to why it has chosen $(-2,-5)^T$, well that's a mystery. Perhaps they have done it so that the diagonalising matrix (usually called $P$) has a positive determinant, rather than a negative one. This will ensure the transformation $P$ represents is not a reflection.