Repeated Eigenvalues — Question 2

Question 2 – Solution

Strategy. A zero eigenvalue can preserve an equilibrium line while a nilpotent part creates unbounded drift parallel to it.

Step 1: Identify the nilpotent part and equilibria. Direct multiplication gives $N^2=0$ but $N\ne 0$. The equation $NX=0$ requires $x=y$, so the equilibria form the one-dimensional line $L=\{(s,s):s\in\mathbb R\}$. The double zero eigenvalue has only one independent eigendirection.

Step 2: Construct the complete flow. The matrix $F(t)=I+tN$ has $F(0)=I$ and $F'=N=NF$, since $N^2=0$. Thus $$\boxed{X(t)=\binom pq+t(p-q)\binom 11.}$$ Also $(I+tN)(I+sN)=I+(t+s)N$, verifying the evolution law and $F(t)^{-1}=F(-t)$. All real initial states and all real times are allowed.

Step 3: Find the drifting straight lines. Put $d=p-q$. If $d\ne 0$, then $x-y=d$ for every time and the velocity is the constant vector $d(1,1)^T$. Thus the full orbit is the entire line $x-y=d$, traversed up/right for $d>0$ and down/left for $d<0$. Its distance to $L$ stays $|d|/\sqrt 2>0$, so it neither reaches nor approaches the equilibrium line. If $d=0$, the orbit is one fixed point, not the entire line $L$.

Step 4: Test boundedness and stability. Exactly $p=q$ gives a forward-bounded solution; all other solutions drift without bound. The origin is unstable: initial $(\varepsilon,0)$ can be arbitrarily small, yet evolves to $(\varepsilon(1+t),\varepsilon t)$, eventually leaving any fixed neighborhood. Zero real parts alone do not ensure boundedness or stability when a nonzero nilpotent part is present. The dashed line in the figure consists of separate equilibria.