I have been learning the basics of optimal control (from here) and have a question about solving a linear ODE. Let's say I have the following system
$$\dot{x} = A x$$
whose solution is
$$x(t) = e^{At} x(0)$$
From what I have been reading, this is difficult to solve, because $e^{At}$ is difficult to evaluate. The only way to evaluate it is to use a Taylor series expansion, but this is difficult to compute in practice.
Therefore, the solution is to represent the state in terms of the eigenvectors of $A$. After doing some algebra, this results in the following solution:
$$x(t) = T e^{Dt} T^{-1} x(0)$$
where $T$ is a matrix of the eigenvectors of $A$, and $D$ is a diagonal matrix of the eigenvalues of $A$.
Apparently, it is easier to now compute $e^{Dt}$ than it was to compute $e^{At}$ because $D$ is diagonal. But I don't understand why this is. Why does computing $e^{At}$ needs the Taylor series expansion, but computing $e^{Dt}$ can be done directly?
Recall that as you say, $\exp(At)$ is defined by the power series $$ \exp(At)=\sum_{n=0}^\infty \frac{t^n}{n!}A^n $$ Do you see why the powers $A^n$ are easier to compute if $A$ is diagonal?