Linear dynamical systems

Question

Show that, if the real system $$\dot{x}=\left(\begin{array}{cc}\alpha&-\beta\\\beta&\alpha\end{array}\right)x$$ is diagonalised over the complex numbers $\mathbb{C}$, such that the transformed coordinates $(z_1,z_2)$ are taken relative to complex conjugate eigendirections, then $z_1=\overline{z_2}=z$. Hence, deduce that the real system is effectively replaced by a single complex differential equation $$\dot{z}=(\alpha+i\beta)z.$$ I guess the second part is not difficult to show, but I cannot deal with the first one even if it seems very simple and natural...

Answer 1

Maybe you are making things more complicated than they are... Denote by $u$ and $v$ the entries of $x$, that is, let $x(t)=\begin{pmatrix}u(t)\\ v(t)\end{pmatrix}$, then the complex valued $z(t)=u(t)+\mathrm iv(t)$ indeed solves $$\dot z(t)=(\alpha+\mathrm i\beta)z(t),$$ hence, for every $t$, $$z(t)=\mathrm e^{(\alpha+\mathrm i\beta)t}z(0),$$ that is, using the definitions of $z(t)$ and $z(0)$, $$u(t)+\mathrm iv(t)=\mathrm e^{\alpha t}(\cos(\beta t)+\mathrm i\sin(\beta t))(u(0)+\mathrm iv(0)).$$ Identifying the real and imaginary parts of the RHS, one gets $$u(t)=\mathrm e^{\alpha t}(\cos(\beta t)u(0)-\sin(\beta t)v(0)),\qquad v(t)=\mathrm e^{\alpha t}(\cos(\beta t)v(0)+\sin(\beta t)u(0)),$$ which can equivalently be written in vector form as $$x(t)=\mathrm e^{\alpha t}\cos(\beta t)\begin{pmatrix}u(0)\\ v(0)\end{pmatrix}+\mathrm e^{\alpha t}\sin(\beta t)\begin{pmatrix}-v(0)\\ u(0)\end{pmatrix},$$ or in full matrix form as $$x(t)=\mathrm e^{\alpha t}\begin{pmatrix}\cos(\beta t)&-\sin(\beta t)\\\sin(\beta t)&\cos(\beta t)\end{pmatrix}x(0).$$

Answer 2

In this post, Complex Numbers vs. Matrix, we see the homeomorphism between complex numbers and matrices of the form $$ \left[ \begin{array}{cr} a & -b \\ b & a \end{array} \right] $$ From that we can immediately state that $$ \dot{z} = z_{0} z $$ where $z_{0} = a + b i$.