I have a linear system of homogeneous ordinary differential equations, i.e.:
$$ \dot{x}=Ax $$
where $A$ is an $n\times n$ real matrix.
The matrix exponential method (described for example here) tells me that
$$ e^{At} C $$
where $C=(C_1,C_2,C_3)$ are arbitrary constants, is the general solution to the system.
There is another method, which I will call here the eigenvectors method, (described in three parts here) that builds the solutions step-by-step, column-by-column. If $A$ is diagonizable with eigenvectors $v_1,\dots,v_n$ we obtain the solution
$$ x=C_1 e^{\lambda_1 t} v_1 + \dots + C_n e^{\lambda_n t} $$
If $A$ is not diagonizable, then we use the Jordan form of $A$: for generalized eigenvectors we have generalized summands. So if $v$ is not an eigenvector, but a generalized eigenvector, then instead of writing $C_i e^{\lambda_i t} v$ in the sum, we write
$$ C_i e^{\lambda_i t}\left( 1+t+\frac{t^2}{2} + \frac{t^3}{3!} + \dots + \frac{t^n}{n!} \right) v\qquad (*)$$
where $n$ is the rank of the generalized eigenvector $v$.
Now comes my question. When I first saw these two methods, I thought they are one and the same method. It's because, to compute $e^{At}$ we also need to compute the (maybe generalized) diagonalization $A=M D M^{-1}$ where $D$ is in Jordan form and $M$ is a base changing matrix. Then, by a known formula
$$ e^{At}= e^{M(Dt)M^{-1}} = M^{-1}e^{Dt}M $$
and computing $e^{Dt}$ can be done easilly. If $D$ is diagonal, then $e^{Dt}$ is just element-wise exponenciation. If there is a off-diagonal nonzero element, we get something resembling $(*)$. So the eigenvectors method is just matrix exponential method in disguise, right?
Sadly no. Even if $D$ is indeed diagonal, then the matrix exponential method will give
$$ x = M^{-1} e^{Dt} M C $$
but the eigenvector method will give instead
$$ x = e^{Dt} M C $$
and this is really confusing for me. Why on Earth would I want to compute $M^{-1}$, when it's completely unnecessary? I just fail to understand why the matrix exponential method exists. So I want to know why. For me now it looks like the matrix exponential does more work for the uglier results, because almost always $e^{Dt}M$ is simple and $M^{-1} e^{Dt} M$ is ugly. Maybe I am wrong or something that I've written above is wrong, that answers my question? (The question is motivated by an 1 hour+ of expanding out $M^{-1} e^{Dt} M$, after when I discovered the more elegant method.)
Best Answer
The eigenvectors method is a consequence of the matrix exponential method. You get (*) by computing $e^{At}v$, where $v$ is a generalized eigenvector. The reason you got two different matrices is that they are two different fundamental matrices for equation $$ \dot{x} = Ax. $$ In other words, both $X_1(t)=e^{At}$ and $X_2(t)$ obtained by the eigenvectors method satisfy this equation, and the general solution is in the form $$ x(t)=X_1(t)C_1 $$ or equivalently $$ x(t)=X_2(t)C_2. $$ However, if you add initial conditions, the constants $C_1$ and $C_2$ will be different.
So to sum up, the eigenvectors method relies on the form of exponential matrix, but indeed it is usually nicer to apply than obtaining the exponential matrix itself.