# Why do we use the method of matrix exponential?

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.)

• $x$ 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... – Quillo May 27 '20 at 18:00
• Right. I added vector of constants $C$. Is it correct now? – mz71 May 27 '20 at 18:05
• 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 $M^{-1}$. – John B May 27 '20 at 18:07
• 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. – amd May 27 '20 at 19:33
• 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. – 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.
• 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. – M_S May 27 '20 at 19:15
• "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! – mz71 May 27 '20 at 20:20