So I’m trying to solve the system $x’=Ax$ with the initial conditon $x(0)=v_4$ with $A$ a 4x4 matrix with constant coeficients. And I am given the following properties;
$$Av_1=2v_1\,;Av_2=-3v_2\,; Av_3=2v_3+v_1\,; Av_4=2v_4+v_3$$
I really don’t know how to tackle this problem, I thought about using the exponential matrix as I want to know $e^{At}v_4$, and it seems like I have to use generalized eigenvectors, but I don’t fully understand them. I’d be very thankful if someone could explain how can I solve this type of exercise!
Assuming $(v_1,v_2,v_3,v_4)$ is a basis of your space you have been given a way to find your matrix $A$ written in the basis $(v_1,v_2,v_3,v_4)$. I will start by giving you the matrix $A$ and work you through its derivation: $$ A=\begin{bmatrix} 2 & 0& 1& 0\\ 0& -3& 0& 0\\ 0& 0& 2& 1\\ 0& 0& 0&2 \end{bmatrix}. $$ The first property is that $Av_1=2v_1$. Thus $v_1$ is clearly an eigenvector to $A$ with eigenvalue 2. Since the matrix is written in the assumed basis, this means that the first column of $A$ has to double the vector. Similarly for the second property.
For the third property you want $Av_3=2v_3+v_1$, i.e. the linear transformation applied to $v_3$ should return two times $v_3$ together with one $v_1$. Similarly for your last property.
As mentioned before this gives you the matrix $A$ written in the basis of $(v_1,v_2,v_3,v_4)$. I assume you know these vectors. Regardless you can rewrite the matrix in respect to the standard basis with a basis change. Thereafter you are free to find the exponential matrix of A and solve your system.
You mention the use of generalized eigenvectors to solve this problem. This is possible rewriting the above matrix to be in Jordan form, but if you're not yet familiar with generalized eigenvectors, i doubt that to be the goal of the exercise.