# System of linear homogeneous differential equations with complex eigenvalue

I need to solve the system of the DE:
$$\begin{cases} \frac{dx}{dt} = 2x-4y, \\ \frac{dy}{dt} =x+2y+z,\\ \frac{dz}{dt} = 3y+2z, \end{cases}$$ First, I rewtite the coefficents in the matrix form: $$A = \begin{bmatrix} 2 & -4 & 0\\ 1 & 2 & 1\\ 0 & 3 & 2 \end{bmatrix}$$
Then, I find $$det(A-\lambda I) = \begin{vmatrix} 2 - \lambda & -4 & 0\\ 1 & 2 - \lambda & 1\\ 0 & 3 & 2 - \lambda \end{vmatrix} =-(\lambda-2)(\lambda-(2-i))(\lambda-(2+i))$$ For $$\lambda =2$$ the approach is not very difficult - I will solve the system $$Ev =0$$ where $$E=A-\lambda I$$ and $$v = \begin{bmatrix} x\\ y\\ z \end{bmatrix}$$.
But this approach is much for difficult more complex $$\lambda$$.

• What does the general solution look like for a $2\times2$ coefficient matrix with complex eigenvalues? Extrapolate from that. – amd Mar 4 at 7:02

If we stay within real matrices, we cannot diagonalize the coefficient matrix, but we can block-diagonalize it.For $$\lambda=2$$, an eigenvector is $$\begin{bmatrix}-1\\0\\1\end{bmatrix}$$ For $$\lambda=2+i$$, an eigenvector is $$\begin{bmatrix}-4\\i\\3\end{bmatrix}=\begin{bmatrix}-4\\0\\3\end{bmatrix}+\begin{bmatrix}0\\1\\0\end{bmatrix}i$$ Let $$P=\begin{bmatrix}-1&-4&0\\0&0&-1\\1&3&0\\\end{bmatrix}$$ Let $$\check D=\begin{bmatrix}2&0&0\\0&2&-1\\0&1&2\end{bmatrix}$$ Then $$P^{-1}AP=\check D$$ Let $$\begin{bmatrix}x\\y\\z\end{bmatrix}=P\begin{bmatrix}u\\v\\w\end{bmatrix}$$ Then $$\begin{bmatrix}du/dt\\dv/dt\\dw/dt\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&-1\\0&1&2\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}$$ $$u=ae^{2t},v=e^{2t}(b \cos t+c \sin t),w=e^{2t}(-c \cos t+b \sin t)$$. You can take it from here.
• Could you please comment oh how you got eigenvector for $\lambda = 2+i$. You just took the second column of $A-\lambda I$ matrix? – student Mar 4 at 14:24