Solution to Differential system with Complex Eigenvectors

Solution 1:

You don’t really need the eigenvectors to generate a solution to this system. The characteristic polynomial of the coefficient matrix $A$ is $\lambda^2-2\lambda+2 = (\lambda -1)^2+1$, therefore $(A-I)^2+I=0$ by Cayley-Hamilton. Set $B=A-I$. Since the identity matrix commutes with everything $e^{tA}=e^{t(I+B)}=e^te^{tB}$. Expand the power series of $e^{tB}$ and use the fact that $B^2=-I$: $$\begin{align} e^{tB} &= I+tB+{t^2\over2!}B^2+{t^3\over3!}B^3+{t^4\over4!}B^4+\cdots \\ &= I + tB-{t^2\over2!}I-{t^3\over3!}B+{t^4\over4!}I+\cdots\end{align}$$ Gathering up terms involving $I$ and $B$, we find that the coefficient of $I$ is the power series for $\cos t$ and the coefficient of $B$ is the power series for $\sin t$. Putting all this together, $$e^{tA}=e^t\cos(t)\,I+e^t\sin(t)\,B = e^t\begin{bmatrix}\cos t-3\sin t & 5\sin t \\ -2\sin t & \cos t + 3\sin t\end{bmatrix}.$$

More generally, you can use a similar calculation to find that if the eigenvalues of a $2\times2$ matrix $A$ are $\alpha\pm i\beta$, then $$e^{tA} = e^{\alpha t}\left(\cos{\beta t} \, I+{\sin{\beta t}\over\beta}B\right),$$ where $B=A-\alpha I$.

That this looks a lot like $e^{\alpha+i\beta}$ is no coincidence. Under matrix multiplication, matrices of the form $\tiny{\begin{bmatrix}\alpha&-\beta\\\beta&\alpha\end{bmatrix}}$ (conformal matrices) are isomorphic to the complex numbers, and a $2\times2$ real matrix with complex eigenvalues is similar to such a matrix.