How find the general solution to a linear differential system with complex coefficients

Find the general solution to the following system:
$$\dot{\textbf{x}}=\begin{pmatrix} 0 & \mathrm{i}\\ \mathrm{-i} & \mathrm{2i} \end{pmatrix}\textbf{x}$$
Let $\textbf{A}=\begin{pmatrix} 0 & \mathrm{i}\\ \mathrm{-i} & \mathrm{2i} \end{pmatrix}$. I want to solve this system by finding the eigenvalues $\lambda_1,\lambda_2$ and the corresponding eigenvectors $\textbf{v}_1, \textbf{v}_2$ of $\textbf{A}$, but when I take a step further to write the solution, I have no idea whether I can write the solution into real and imaginary parts and omit the imaginary unit $\mathrm{i}$ of the imaginary part as we often do for linear differetial system with real coefficients.
If I can't, is it correct to reserve $\mathrm{i}$? Or what should I do?
Any help appreciated!

You cannot take the real part.

Your matrix is not diagonalizable, so you have to look into the generalized eigenspaces. But in this case $(A-iI)^2=0$, which simplifies things a lot.

The solution of the system is $\mathbf x(t)=e^{At}\mathbf x_0$. In this particular case, since $(A-i)^2=0$, we have $$e^{(A-i)t}=I+(A-iI)t,$$ so $$e^{At}=e^{(A-i)t}e^{it}=e^{it}\,I+t\,e^{it}(A-iI)=(1-it)e^{it}\,I+te^{it}A =\begin{bmatrix}(1-it)e^{it}&ite^{it}\\-ite^{it}&(1+it)e^{it}\end{bmatrix}.$$

So now you write $\mathbf x_0=\begin{bmatrix}c_1\\ c_2\end{bmatrix}$, and the solution becomes $$\mathbf x(t)=e^{At}\mathbf x_0=\begin{bmatrix}(1-it)e^{it}&ite^{it}\\-ite^{it}&(1+it)e^{it}\end{bmatrix}\begin{bmatrix}c_1\\ c_2\end{bmatrix}=\begin{bmatrix} c_1(1-it)e^{it}+c_2ite^{it}\\ -c_1ite^{it}+c_2(1+it)e^{it}. \end{bmatrix}$$

• I'm curious how you come up with $(\textbf{A}-\mathrm{i}\textbf{I})^2=0$. It is always time-consuming to solve the eigenvalues and eigenvectors of a matrix. I'd like to know some shortcut in solving problems like this (not necessarily with complex coefficients). My professor always gives us dozens of problems in exam :( Thanks a lot! – R.C Dec 2 '17 at 5:36
• If you don't have enough eigenvectors (that is, if the matrix is not diagonalizable), after you find the eigenvectors you need to look into the generalized eigenvectors. The first step is the kernel of $(A-\lambda I)^2$. But, in the $2\times2$ case, this is always zero. – Martin Argerami Dec 2 '17 at 6:30

There are no nonzero real-valued solutions because the differential equation would imply that the derivative of such a solution is nonreal. You should leave your answer as a complex-valued function.

Bear in mind that this matrix is not diagonalizable, so finding eigenvectors and eigenvalues alone is not enough.

• Does it mean that if the coefficient matrix has complex numbers, the solution is always not real, and we need to keep the imaginary part? – R.C Dec 2 '17 at 3:50
• @R.C You should always consider the imaginary part... for some real matrices, you can combine the real and imaginary parts of eigenvectors to produce real-valued solutions – user16394 Dec 2 '17 at 4:41