General solution of $X' = ((0,0),(0,0))X$

Question

So usually we use the $\det(A-\lambda I)$ to get eigenvalues to get eigenvectors to solve for a general solution. This seems neither necessary nor helpful here, just wondering if the general solution is just $(c_1, c_2) = X$ where $c_1$ and $c_2$ are constants in complex numbers, since the derivative matrix is always 0.

Thanks!

Answer 1

You can execute the matrix-vector multiplication: $$X'(t)=\left(\begin{matrix} 0&0\\ 0&0 \end{matrix}\right)X(t)$$ $$X'(t)=\left(\begin{matrix} 0\\ 0 \end{matrix}\right)$$ $$\left(\begin{matrix} x_1'(t)\\ x_2'(t) \end{matrix}\right)=\left(\begin{matrix} 0\\ 0 \end{matrix}\right)$$ So your equations are: $$x_1'(t)=0$$ $$x_2'(t)=0$$ And I think you can solve them.

Answer 2

Note that:

$$ \int_0^t X'(\tau)d\tau = X(t)-X(0) $$

but due to your equation:

$$ \int_0^t X'(\tau)d\tau = \int_0^t \left(\begin{matrix} 0&0\\ 0&0 \end{matrix}\right) X(\tau)d\tau = \left(\begin{matrix} 0&0\\ 0&0 \end{matrix}\right) \int_0^t X(\tau)d\tau = \left(\begin{matrix} 0\\ 0 \end{matrix}\right) $$

Hence $X(t)$ is a constant.