Let $A$ be a 3 by 3 matrix, such that $\dot{x}=Ax$. I am trying to find the fundamental matrix solution. I know that I need to find the eigenvalues and eigenvectors of $A$ which I did but I am not sure what to do next. Does anyone know what to once you find the eigenvalues and eigenvectors?
Thanks in advance.
Assuming you have three unique eigenvalues $(\lambda_1,\lambda_2,\lambda_3)$ and eigenvectors $(\pmb{\xi}^{(1)},\pmb{\xi}^{(2)},\pmb{\xi}^{(3)})$ you should have three linearly independent solutions in the form $$\mathbf{x}_1(t)=\pmb{\xi}^{(1)} e^{\lambda_1t}\qquad \mathbf{x}_2(t)=\pmb{\xi}^{(2)} e^{\lambda_2t}\qquad \mathbf{x}_3(t)=\pmb{\xi}^{(3)} e^{\lambda_3t}$$ Then your fundamental matrix should be $$\pmb{\psi}(t)=\left( \begin{array}{@{}ccc@{}} \mathbf{x}_1(t)& \mathbf{x}_2(t)& \mathbf{x}_3(t) \end{array}\right)=\left( \begin{array}{@{}ccc@{}} \xi_1^{(1)e^{\lambda_1t}}&\xi_1^{(2)e^{\lambda_2t}}&\xi_1^{(3)e^{\lambda_3t}}\\ \xi_2^{(1)e^{\lambda_1t}}&\xi_2^{(2)e^{\lambda_2t}}&\xi_2^{(3)e^{\lambda_3t}}\\ \xi_3^{(1)e^{\lambda_1t}}&\xi_3^{(2)e^{\lambda_2t}}&\xi_3^{(3)e^{\lambda_3t}}\\ \end{array} \right) $$ where $\xi_n^{(m)}$ denotes the $n$th element of $\pmb{\xi}^{(m)}$. Note that the general solution of the differential equation is $$\mathbf{x}=\pmb{\psi}(t)\mathbf{c}$$ where $\mathbf{c}=(c_1,c_2,c_3)^\intercal$ is a constant matrix.