Cayley-Hamilton theorem illustration
CAYLEY-HALMITON theorem
It states that any square 
matrix 
is solution of its characteristic equation. That is to say: let us denote 
, the vector of the n characteristic polynomial coefficients such that:
matrix is solution of its characteristic equation. That is to say: let us denote , the vector of the n characteristic polynomial coefficients such that:
Then:
Computation of 
from 
, 
, 
, ⋯, 
, 
:
Then by using this theoreom it can be shown that, for 
:
:
where the coefficients 
are recursively defined by:
are recursively defined by:
, 
and for 
,
,
,
.
Notation : 
.
.Illustration :
n=5;
A=rand(n,n);
a=poly(eig(A));
vector 
:
:a=a(n+1:-1:2);
vector 
:
:f0=-a;
exemple: 
(the objective is to compute 
):
(the objective is to compute ):k=3;
Recurcive computation of 
:
:fk=f0;
for ii=1:k,
temp=-a(1)*fk(end);
fk(2:end)=-a(2:end)*fk(end)+fk(1:end-1);
fk(1)=temp;
end
Check the result:
bid=zeros(n,n);
for ii=1:n, bid=bid+fk(ii)*A^(ii-1);end;
bid-A^(n+k)
Computation of 
from , 
, 
, ⋯, 
, :
The characteristc polynomial of 
is:
is:
Then, for :
:
where the coefficients 
are recursively defined by:
are recursively defined by:
,
and for ,
,
,
.
Notation : 
.
.Computation of 
from , , , ⋯, :
with:
Notation : 
.
.Remark: in the following example we show how the matrix exponential can be computed thanks to the Cayley-Hamilton theorem. It is just an illustrative example. There is more efficient algorithms to compute the matrix exponential (see: help expm ).
Example (in the sum operation, k is limited to 
):
):t=0.1;
At=A*t;
Vector of characteristic polynomial coefficients:
for ii=1:n, at(ii)=a(ii)*t^(n+1-ii);end;
f0t=-at;
Recursive computation of 
:
:gt=1./factorial([0:n-1]);
for k=0:20,
fkt=f0t;
for ii=1:k,
temp=-at(1)*fkt(end);
fkt(2:end)=-at(2:end)*fkt(end)+fkt(1:end-1);
fkt(1)=temp;
end
gt=gt+1/factorial(n+k)*fkt;
end
Check the result:
bid=zeros(n,n);
for ii=1:n, bid=bid+gt(ii)*A^(ii-1)*t^(ii-1);end;
bid-expm(At)