Introduction and Background

In the scalar case, the solution to the differential equation u'=au is given by u=CExp[a t]. By analogy, we will investigate what happens in the case of a system u'=Au, where A is a square constant coefficient matrix.

Look for exponential solutions u=Exp[ct]x (here x is a vector). The differential equation becomes
c Exp[ct]x=A Exp[ct]x.
This implies that c is an eigenvalue of A.

Let's look at this another way. If A is diagonal, the system of differential equations is "uncoupled", that is, the components do not depend on each other.

In[1]:=

  MatrixForm[{dx1,dx2}]
  MatrixForm[{{2,0},{0,3}}]
  MatrixForm[{{2,0},{0,3}}.{x1,x2}]

Out[1]=

  dx1
  
  dx2

Out[2]=

  2   0
  
  0   3

Out[3]=

  2 x1
  
  3 x2

Looking at this we see that
dx1=2x1
dx2=3x2
(treat the d as d/dx).
These equations are uncoupled and the solutions are exponential functions.

We can now see how a solution to u'=Au can be written as u=S.Diagonal[{Exp[lambda1 t], . . . Exp[lambdan t}]Inverse[S]. Recall that if A is diagonalizable, the diagonal matrix is Inverse[S].A.S, where S is the matrix of eigenvectors. It is the diagonal matrix we want to look at more closely.

Recall that Exp[t]=1+t+t^2/2!+t^3/3!+ . . .

We define the matrix exponential function as Exp[At]=I+At+(At)^2/2!+(At)^3/3!+ . . .

This definition keeps most of the important properties of the exponential function, some suitably modified:
1. Exp[0 matrix]=Identity Matrix
2. Exp[t(A+B)]=Exp[tA].Exp[tB] if A.B=B.A (A and B commute)
3. Exp[(t1+t2)A]=Exp[t1A].Exp[t2A]
4. d/dt(Exp[tA].c)=Exp[tA].A.c
5. Inverse[Exp[tA]]=Exp[-tA]

Up to Matrix Exponentials and Differential Equations