Given the vector equation $$x' = Ax+b$$ where $x,b \in \mathbb{R}^n, x(0) = x_0, A\in \mathbb{R}^{n\times n},x'=\frac{dx}{dt}$ and $A$ is self-adjoint, how does the solution to such an equation change if $A$ has $0$ as an eigenvalue i.e. it is a singular matrix?
I know if your matrix can be decomposed into a diagonalizable form, that is $$A=Q\Lambda Q^{-1}$$ the solution to the above equation is given by
$$x=Qe^{\Lambda t}(Q^{-1}x_0 + \Lambda ^{-1} Q^{-1}b)-Q\Lambda Q^{-1}b$$ Is the difference when $A$ is singular that instead of a nice diagonal matrix of eigenvalues of $\Lambda$, I will have to use the Jordan form to solve the problem? Can I use the Jordan form to find $e^{At}$?