I have subin the eigenvalue then I got the matrix
[0 1 3 1]
[0 1 2 4]
[0 0 2 2]
[0 0 1 3]
reduce to
[0 1 0 -8]
[0 1 0 2]
[0 0 2 2]
[0 0 1 3] does this imply x3 = x4 =0? what is x1 then, how do I write it in terms of t and s?
You have "subin"...?? Anyway, if the last matrix is the reduced one to find the eigenspace of whatever, then indeed the 3rd and 4th rows imply that $\displaystyle x_3=x_4=0$ , and the 2nd row then implies $\displaystyle x_2=0$ as well...and $\displaystyle x_1$ can be freely chosen , so in this case the subspace (eigenspace or whatever) is $\displaystyle \left\{\begin{pmatrix}x\\0\\0\\0\end{pmatrix}\,,\, \,any\,\,\,x\right\}$.
Tonio