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.)

  • $\begingroup$ $x$ is a vector, $M$, $M^{-1}$ and $e^{Dt}$ are matrices. I do not understand.. $x = M^{-1} e^{Dt} M$ is "vector = matrix". Something is wrong, or you forgot some symbols... $\endgroup$ – Quillo May 27 '20 at 18:00
  • $\begingroup$ Right. I added vector of constants $C$. Is it correct now? $\endgroup$ – mz71 May 27 '20 at 18:05
  • $\begingroup$ The two give the same. In your equation $(*)$ you are assuming that the matrix is already in Jordan form (or some block of it). To bring it back to the "usual" form you need to multiply by $M^{-1}$. $\endgroup$ – John B May 27 '20 at 18:07
  • 1
    $\begingroup$ You don’t always need to diagonalize in order to compute the exponential of a matrix. See this for simple ways to compute the exponential of any $2\times2$ matrix without computing any eigenvectors whatsoever. Some of those special cases carry over to larger systems. You’ve also probably only worked with artifically-constructed “nice” matrices for which the eigenvectors and eigenvalues are simple and easily-computed. That’s not true in general, and the power series for $e^{tA}$ gives you a way to get a numerical approximation. $\endgroup$ – amd May 27 '20 at 19:33
  • 1
    $\begingroup$ Also, once you start working with inhomogeneous equations, the variation of parameters method for finding a particular solution is quite straightforward in terms of the exponential of the coefficient matrix. $\endgroup$ – amd May 27 '20 at 19:34

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.

  • $\begingroup$ Yes, but my question is why the matrix exponential could be useful. Why don't we use it to prove that it works, and then forget about it entirely? $\endgroup$ – mz71 May 27 '20 at 18:49
  • 1
    $\begingroup$ Honestly I now can't think of an example where calculating the exponential matrix would have some big advantage compared to the other method. I suppose it could be more useful if you add the initial condition $x(0)=x_0$, then the solution is just $$x(t) = e^{At}x_0, $$ where for the other fundamental matrix you have $X(t)X^{-1}(0)x_0$, so either way you need to invert some matrix. However exponential matrix is imo better to understand the qualitative properties of solutions, for example if the solution tends to $0$ or not. $\endgroup$ – M_S May 27 '20 at 19:15
  • $\begingroup$ "However exponential matrix is imo better to understand (...) if the solution tends to $0$ or not." - exactly what I wanted to hear. If there is anything the exponential can do better, I want to hear it! $\endgroup$ – mz71 May 27 '20 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.