Complex Eigenvalues — Question 1

Question 1 – Solution

Strategy. Take real and imaginary parts with their signs intact, then use polar coordinates to separate decay from rotation.

Step 1: Extract real solutions. The eigenvalues are $-1\pm 2i$. For $\lambda=-1+2i$, choose $v=(1,-i)^T$, which satisfies $Av=\lambda v$. Writing $e^{\lambda t}v=U+iV$ gives $$U=e^{-t}\binom{\cos 2t}{\sin 2t},\qquad V=e^{-t}\binom{\sin 2t}{-\cos 2t}.$$ Because $A$ is real, both parts solve the original real system.

Step 2: Resolve the initial state. The determinant of the columns $(U,V)$ is $-e^{-2t}\ne 0$. At zero they are $(1,0)^T$ and $(0,-1)^T$, so $X=U-2V$: $$\boxed{X=e^{-t}\binom{\cos 2t-2\sin 2t}{\sin 2t+2\cos 2t}.}$$ The nonzero determinant and linear-system uniqueness give the complete real family as arbitrary real combinations of $U,V$.

Step 3: Locate the first positive-axis crossing. Let $\theta_0=\arctan 2$, which lies in $(0,\pi/2)$. Here $r=\sqrt 5e^{-t}$ and a continuous angle is $\theta=\theta_0+2t$. Rotation is counterclockwise. The first positive time on the positive $x$-axis requires $\theta=2\pi$, hence $\boxed{t_*=(2\pi-\arctan 2)/2}$. The earlier negative-axis crossing is not the requested event.

Step 4: Separate revolution from repetition. One revolution takes $\pi$, reducing the radius by $e^{-\pi}$. Since the radius strictly decreases, no nonzero solution is periodic; the origin is a stable spiral. Conjugating the exponential solution gives $U-iV$, whose parts are $U,-V$, already in the same real span. Thus a conjugate pair supplies two real degrees of freedom, not four. The figure shows a finite forward segment of the stated IVP.