I am trying to solve the following system. \begin{aligned} x'' &= -x + y\\ y'' &= x - y \\ \end{aligned} I reduce this to a first order system, \begin{aligned} x' &= x_1 \\ y' &= y_1 \\ x_1' &= -x + y \\ y_1' &= x - y \\ \end{aligned}
The eigenvalues are $0$, $0$, $i\sqrt{2}$ and $-i\sqrt{2}$.
I call the coefficient matrix $$ {\bf A} = \begin{pmatrix} 0&0&1&0 \\ 0&0&0&1 \\ -1&1&0&0 \\ 1&-1&0&0 \end{pmatrix} $$ Now when I solve for ${\bf A}{\bf V}_1 = {\bf 0}$ for the $0$ eigenvalue, I get an eigenvector ${\bf V}_1 = (1,1,0,0)^\top$. When I solve for ${\bf A}{\bf V}_2 = {\bf V}_1$, I get the following equations \begin{aligned} &v_{21} - v_{22} = -1 \\ &v_{21}-v_{22} = 1 \\ &(v_{23}=v_{24}=0) \end{aligned} which do not have any solution.
P.S. I know how to solve this system using for $0$ eigenvalue with $(a + bt){\bf V}_0$ with ${\bf V}_0$ being eigenvector for the $0$ eigenvalue. I want to know what is wrong with the first-order approach.
You got your equations wrong. Solving $$ \begin{pmatrix} 0&0&1&0 \\ 0&0&0&1 \\ -1&1&0&0 \\ 1&-1&0&0 \end{pmatrix} \pmatrix{v_{21}\\v_{22}\\v_{23}\\v_{24}} = \pmatrix{1\\1\\0\\0} $$ is equivalent to the equations $$ v_{23}=1\\ v_{24}=1\\ -v_{21}+v_{22}=0\\ v_{21}-v_{22}=0 $$ which has solutions $\pmatrix{c\\c\\1\\1}$, of which one can select for instance $$ V_2=\pmatrix{0\\0\\1\\1}. $$