Solving coupled ODE system with one eigenvalue

I have the coupled system:

$$\left\{ \begin{array}{c} dx/ds=p_1 \\ dp_1/ds=-x \end{array} \right.$$

I Tried solving it with the change of coordinates (eigenvalue, eigenvector) approach, but I ended up with one eigenvalue: $$\lambda = i$$ Therefore I do not know how to form my decoupling matrix, because AFAIK there are 2 methods for building it: finding the inverse matrix of the eigenvector matrix or making a diagonal matrix with the eigenvalues on the main diagonal. Since there is only one eigenvalue, the eigenvector matrix is not square and it has no inverse matrix and using the other method just produces a single number: i, which would not decouple the system.

Is there any other approach for building the decoupling matrix?

If there is not, what other method could I use, to solve the initial problem?

Thank you!

• You can still diagonalize this system using both the eigenvalue $i$ and the eigenvalue $-i$. These each have one linearly independent eigenvector so you can proceed in the usual way (though you will need to deal with complex number arithmetic). You can save some calculation by noticing that if $A$ is real $v$ is an eigenvector $\lambda$ then $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda}$.
– Ian
Commented Jun 15, 2017 at 13:58
• This of course does not work as given for a matrix that is actually nondiagonalizable like $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$; in these cases you need to know the exponential of larger Jordan blocks.
– Ian
Commented Jun 15, 2017 at 14:01

So, @Ian's comment made me do my calculation once again, which led me to the conclusion, that I was wrong before and the system actually has 2 eigenvalues $$\lambda_1 = 1 \ \ and\ \lambda_2 = -1$$ Therefore, I can use the usual algorithm and the impediment is gone.