I am trying to understand how to find eigenvalues from a matrix consisting of exponential terms, considering a differential equation. The examples I've seen online are ODEs. Without using a vector with exponential terms, here is what I have learned.
$$\dfrac{d}{d t}\overrightarrow{x}(t)=\lambda \overrightarrow{x}(t)$$
$$\dfrac{d}{d t} \begin{bmatrix} \ x_1(t) \\ \ x_2(t) \\ \ x_3(t) \end{bmatrix}= \begin{bmatrix} \ \lambda_1 & 0 & 0 \\ \ 0 & \lambda_2 & 0 \\ \ 0 & 0 & \lambda_3 \end{bmatrix} \begin{bmatrix} \ x_1(t=0) \\ \ x_2(t=0) \\ \ x_3(t=0) \end{bmatrix}$$
Here is what I am trying to understand
Section of a paper with imag eigenvalue
In this paper, the following assumption is made.
$$\overrightarrow{J}(t)= \lvert \overrightarrow J \rvert e^{-i\omega t}= \begin{bmatrix} \lvert J_x \rvert e^{-i\omega t} \\ \lvert J_y \rvert e^{-i\omega t} \\ \lvert J_z \rvert e^{-i\omega t} \end{bmatrix} $$
They are using partial derivatives (I believe this can be viewed as an ODE then?). Differentiating with respect to time I believe yields the following. Please correct me if I am wrong.
$$\dfrac{\partial}{\partial t}\overrightarrow{J}(t) =\dfrac{\partial}{\partial t} \begin{bmatrix} \lvert J_x \rvert e^{-i\omega t} \\ \lvert J_y \rvert e^{-i\omega t} \\ \lvert J_z \rvert e^{-i\omega t} \end{bmatrix}= \begin{bmatrix} \ -i\omega & 0 & 0 \\ \ 0 & -i\omega & 0 \\ \ 0 & 0 & -i\omega \end{bmatrix} \begin{bmatrix} \lvert J_x \rvert e^{-i\omega t(t=0)} \\ \lvert J_y \rvert e^{-i\omega t(t=0)} \\ \lvert J_z \rvert e^{-i\omega t(t=0)} \end{bmatrix}$$
Or
$$\dfrac{\partial}{\partial t} \begin{bmatrix} \lvert J_x \rvert e^{-i\omega t} \\ \lvert J_y \rvert e^{-i\omega t} \\ \lvert J_z \rvert e^{-i\omega t} \end{bmatrix}= \begin{bmatrix} \ -i\omega & 0 & 0 \\ \ 0 & -i\omega & 0 \\ \ 0 & 0 & -i\omega \end{bmatrix} \begin{bmatrix} \lvert J_x \rvert \\ \lvert J_y \rvert \\ \lvert J_z \rvert \end{bmatrix}$$
Are my assumptions correct? If so, is there a deeper analysis to why this is the case with exponential terms?
Not sure this is what you want to know, but here is a short explanation about the relation between a system of linear ODEs and Eigendecomposition.
Assume a system of first order linear ODEs with constant coefficients. It can be written in matrix form: $$\frac{d\vec x}{dt}=M\vec x.$$
Now assume an exponential solution of the form $$\vec x=\vec ve^{zt}.$$
By plugging it in the equation, you get $$z\vec ve^{zt}=M\vec ve^{zt}$$ which expresses that $z$ is an Eigenvalue of $M$, associated to the Eigenvector $\vec v$: $$M\vec v=z\vec v.$$
So, after finding a diagonalization
$$M=T\Lambda T^{-1}$$ and setting $$\vec x=T\vec y$$ the system reduces to $$\frac{d\vec y}{dt}=\Lambda\vec y$$ which is a system of three easy independent equations
$$\frac{dy_k}{dt}=z_ky_k.$$ The initial conditions can be dealt with by using $\vec y_0=T^{-1}\vec x_0$.