Repeated Eigenvalues — Question 6

Question 6 – Solution

Strategy. A longer nilpotent chain truncates the exponential after a quadratic term; its degree matters for sharp decay estimates.

Step 1: Truncate the nilpotent exponential. Here $N^2$ has only entry $(1,3)$ equal to one; $N^3=0$. Direct differentiation verifies $F(t)=e^{-t}(I+tN+t^2N^2/2)$ and $F(0)=I$. Consequently $$\boxed{X=e^{-t}\begin{pmatrix}p+qt+rt^2/2\\q+rt\\r\end{pmatrix}.}$$ This gives all initial states and shows why a length-three chain needs $t^2/2$, rather than another unrelated eigenvector.

Step 2: Locate the component peaks. For $(p,q,r)=(0,0,1)$, the first components are $t^2e^{-t}/2$ and $te^{-t}$. Their derivatives are $e^{-t}t(2-t)/2$ and $e^{-t}(1-t)$. Thus their maxima are $\boxed{2e^{-2}\text{ at }t=2}$ and $\boxed{e^{-1}\text{ at }t=1}$, respectively. These peaks are not simultaneous.

Step 3: Classify polynomial degrees. If $r\ne 0$, the leading vector factor is $(r/2)t^2(1,0,0)^T$. If $r=0$ but $q\ne 0$, it is $qt(1,0,0)^T$. If $q=r=0$ and $p\ne 0$, it is the constant $p(1,0,0)^T$. Every polynomial factor times $e^{-t}$ tends to zero, so all solutions decay; algebraic multiplicity three does not force degree two for every state.

Step 4: Prove an honest exponential estimate. For initial $(0,0,1)$, $e^t\|X(t)\|\ge t^2/2$, ruling out the proposed constant $C$. The given operator bounds instead imply $\|X(t)\|\le e^{-t}(1+t+t^2/2)\|X(0)\|$. Using $e^{-t/2}\le 1$, $te^{-t/2}\le 2/e$ and $(t^2/2)e^{-t/2}\le 8/e^2$ for $t\ge 0$ gives $$\boxed{\|X(t)\|\le(1+2/e+8/e^2)e^{-t/2}\|X(0)\|.}$$ A slightly slower exponential absorbs the polynomial transient.