Complex Eigenvalues — Question 6

Question 6 – Solution

Strategy. Remove the real part of the spectrum; the remaining matrix acts algebraically like multiplication by an imaginary number.

Step 1: Isolate the oscillatory matrix. The characteristic polynomial is $\lambda^2-2\lambda+5$, with roots $1\pm 2i$. Here $B=\begin{pmatrix}1&-5\\1&-1\end{pmatrix}$ and direct multiplication gives $\boxed{B^2=-4I}$.

Step 2: Construct a real evolution formula. Set $$\boxed{F(t)=e^t\left(I\cos 2t+\tfrac 12B\sin 2t\right).}$$ Its value at zero is $I$. Differentiating the bracket gives $-2I\sin 2t+B\cos 2t$, which equals $B$ times that bracket because $B^2=-4I$. The product rule therefore yields $F'=(I+B)F=AF$. Uniqueness verifies that this formula evolves every initial state.

Step 3: Solve and inspect one oscillation. Since $B(1,1)^T=(-4,0)^T$, $$\boxed{X=e^t\binom{\cos 2t-2\sin 2t}{\cos 2t}.}$$ The initial state is $(1,1)^T$ and initial derivative $(-3,1)^T=A(1,1)^T$. Also $F(\pi)=e^\pi I$: a full trigonometric cycle multiplies the state by $e^\pi$. More generally the periodic bracket and its inverse are bounded, so every nonzero solution grows without bound forward in time. Thus there is no nonzero periodic solution.

Step 4: Verify the evolution algebra. Multiplying two brackets and using $B^2=-4I$ gives the sine and cosine addition formulas, hence $F(t+s)=F(t)F(s)$. Therefore $\boxed{F(t)^{-1}=F(-t)}$. For scalars $c,s$, $\det(cI+sB/2)=c^2+s^2$, so $\boxed{\det F(t)=e^{2t}}$. This controls area, not every vector length: at the state $(1,1)$ the squared-length derivative is $2(1,1)\cdot(-3,1)=-4<0$, despite increasing area. Shape distortion and eventual growth can coexist.