If X' = AX + b then can I use the same principle as if this was an first order ODE?
What do I mean by that, to solve a first order ODE we multiply with the so-called integrating factor: $e^{-\int a(t)dt}$ can I apply this here and solve it this way?
So what would I do:
$$X'-AX=b$$
multiply both sides with $e^{-At}$
$$(Xe^{-At})'=be^{-At}$$
integrate both sides
$$Xe^{-At}=\int be^{-At}dt$$
Giving us:
$$X=e^{At}\int be^{-At}dt$$
Where $A \in \mathcal M_n(\mathbb{R})$
Is this proof correct? How do I integrate a Matrix or a vector? Do I integrate on every component? Also how do I raise $e$ to some matrix power? Do I use the fact that:
$$e^{At}=\sum_{n=0}^{\infty}\frac{(At)^n}{n!}$$
And if $A$ is diagonalizable I can:
$$e^{At}=P(\sum_{n=0}^{\infty}\frac{(Dt)^n}{n!})P^{-1}$$
Now:
$$e^{At}=Pe^{Dt}P^{-1}$$
where: $D$ is a diagonal matrix, but still how do I compute $e^{Dt}$?
And what if $A$ is not diagnolaziable and it's root are complex, conjugate?
How can I solve then?
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain \begin{equation} e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}. \end{equation} Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1. \begin{equation} e^{tD} = \sum_{k=0}^{\infty} \frac{(tD)^k}{k!} = \sum_{k=0}^{\infty} \frac{t^k}{k!} D^k = \sum_{k=0}^{\infty} \frac{t^k}{k!} \left( \begin{array}{cccc} d_1^k\\ & d_2^k\\ & & \ddots\\ & & & d_n^k \end{array} \right) = \left( \begin{array}{cccc} \sum_{k=0}^{\infty} \frac{(td_1)^k}{k!}\\ & \sum_{k=0}^{\infty} \frac{(td_2)^k}{k!}\\ & & \ddots\\ & & & \sum_{k=0}^{\infty} \frac{(td_n)^k}{k!} \end{array} \right) = \left( \begin{array}{cccc} e^{td_1}\\ & e^{td_2}\\ & & \ddots\\ & & & e^{td_n} \end{array} \right). \end{equation} 2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k \geq m$, for some $m \in \mathbb{N}$, you simply obtain a finite sum of powers of $tN$, \begin{equation} e^{tN} = \sum_{k=0}^{m-1} \frac{(tN)^k}{k!}. \end{equation} Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.