Repeated Eigenvalues — Question 1

Question 1 – Solution

Strategy. The eigenspace dimension decides whether a repeated root already supplies a full basis or needs another solution.

Step 1: Distinguish the eigenspaces. For $A$, every nonzero vector is a $-2$ eigenvector, so the eigenspace has dimension two. For $B$, $(B+2I)(x,y)^T=(y,0)^T$, so its eigenspace is $\operatorname{span}(1,0)^T$, of dimension one. Algebraic multiplicity is two in both cases; the geometric multiplicities differ.

Step 2: Recover both complete families. For initial $(p,q)$, $$\boxed{X_A=e^{-2t}\binom pq,\qquad X_B=e^{-2t}\binom{p+qt}{q}.}$$ For $B$, first solve $y'=-2y$, then $(e^{2t}x)'=q$. This derives every solution, and direct differentiation verifies the $qt$ term. In particular $te^{-2t}(1,0)^T$ alone is not the missing solution; the accompanying $e^{-2t}(0,1)^T$ is essential.

Step 3: Analyze the selected orbit. For $B$ with $(p,q)=(0,1)$, $x=te^{-2t}$ and $y=e^{-2t}>0$. Eliminating $t$ gives $$\boxed{x=-\tfrac 12y\ln y,\quad y>0.}$$ Forward time decreases $y$. On $t\ge 0$, $x'=e^{-2t}(1-2t)$ changes sign at $t=1/2$, giving $x_{\max}=1/(2e)$ at $y=e^{-1}$.

Step 4: Interpret multiplicity and geometry. Every nonzero $A$ trajectory is a straight ray toward zero. For $B$, only the eigenline $y=0$ gives straight rays; otherwise $x/y=p/q+t$ changes, giving curved trajectories. All solutions of both systems decay. A repeated root needs no polynomial factor when there is a full eigenbasis; even in the defective system, eigenline initial data have $q=0$ and no $t$ term. The figure shows the selected $B$ orbit, not a generic ray.