Suppose we have 
, 
, and if
, then 
. The same argument will show that 
.
Suppose 
is an analytic function of 
,
then
If 
, 
is a polynomial of degree 
. Then 
and 
.
Example 
.
So 
, 
,
.
Method 2 If 
is simple
If is not simple, e.g., if 
is a double eigenvalue, then
so
If 
,
then 
,
and 
, so
Vandermonde Determinant
One can show
if 's are distinct.
Example 
.
Checks for 
:
for 
;
.
Suppose we are given a matrix
then, to check if 
is equal to for some , we verify
Does 
?
Therefore
If
then
Given 
with I.C. 
, then the
solution is 
. If we differentiate, we get
, so the solution is unique.
Using the natural basis 
of 
as initial conditions. Then let 
be the
solution to 
with 
, then
where is a column vector, and
Then 
and 
.
so
Example 
.
Numerical computation of using Matlab:
with ,
over 
.
, 
.
.
; 
; on
the interval , .
lsim
; before invoking
need to define 
, 
, and 
.
where 
is 
.
Example 
.
In solving for we can make use of block diagonal matrices, for
example
then
where 
, and
, is an easier
problem to solve.
Given 
, 
, if
,
likewise if 
. So the exponential matrix is always invertible,
and 
.
Note:
