Scribd
Upload
40 views2 pages

Putzer Algorithm

The document outlines the Putzer Algorithm for calculating the matrix exponential eAt for a constant n × n matrix A. It defines the algorithm using eigenvalues of A and provides a proof of its uniqueness as a solution to the initial value problem. The proof involves constructing a function Φ(t) and demonstrating its properties using the Cayley-Hamilton theorem.

Uploaded by

nitinvishwakarma6161
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
40 views2 pages

Putzer Algorithm

The document outlines the Putzer Algorithm for calculating the matrix exponential eAt for a constant n × n matrix A. It defines the algorithm using eigenvalues of A and provides a proof of its uniqueness as a solution to the initial value problem. The proof involves constructing a function Φ(t) and demonstrating its properties using the Cayley-Hamilton theorem.

Uploaded by

nitinvishwakarma6161
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 2

Putzer Algorithm to calculate eAt

Matrix Exponential function: Let A be an n × n constant matrix. Then we define the


matrix exponential function by eAt is the solution of the IVP

X ′ (t) = AX(t), X(0) = In .

Putzer Algorithm for finding eAt : Let λ1 , λ2 , · · · , λn be ‘n’ eigenvalues of A. Then


n−1
X
At
e = pk+1 (t)Mk ,
k=0

where M0 = In ,
Mk = Πkr=1 (A − λr In ), 1 ≤ k ≤ n − 1,
and the vector function p(t) = [p1 (t), p2 (t), · · · , pn (t)]t is the solution of IVP,
 
λ1 0 0 ··· ··· 0  
 1 λ2 0

··· ··· 0 
 1
 0 
 0 1 λ3 ··· ··· 0 
 
′
p (t) =   p(t), p(0) =  ..  .
 
 ··· ··· ··· ··· ··· ···   . 
 0 0 0 · · · λn−1 0 
 
0
0 0 0 ··· 1 λn

Proof of the algorithm: Let us define Φ(t) as


n−1
X
Φ(t) = pk+1 (t)Mk , t ∈ R.
k=0

By uniqueness theorem, we are required to prove that Φ(t) is the unique solution of IVP
X ′ = AX, X(0) = In .
n−1
X
Φ(0) = pk+1 (0)Mk = M0 = In , p1 (0) = 1, pk (0) = 0, 2 ≤ k ≤ n.
k=0

As per definition,
′ ′
p1 (t) = λ1 p1 (t), pj (t) = pj−1 (t) + λj pj (t), 2 ≤ j ≤ n.

Now consider for t ∈ R,


n−1 n−1
′
X X
′
Φ (t) − AΦ(t) = pk+1 (t)Mk − A pk+1 (t)Mk
k=0 k=0

n−1 n−1
′
X X
= λ1 p1 (t)M0 + pk+1 (t)Mk − pk+1 (t)AMk
k=1 k=0
n−1
X n−1
X
= λ1 p1 (t)M0 + (λk+1 pk+1 (t) + pk (t))Mk − pk+1 (t)AMk .
k=1 k=0

1
Now
Mk = Πkr=1 (A − λr In ) ⇒ Mk+1 = (A − λk+1 In )Mk = AMk − λk+1 Mk
⇒ AMk = Mk+1 + λk+1 Mk .
Hence from above relation, we find
n−1
X n−1
X
′
Φ (t) − AΦ(t) = λ1 p1 (t)M0 + (λk+1 pk+1 (t) + pk (t))Mk − pk+1 (t)(Mk+1 + λk+1 Mk )
k=1 k=0

n−1
X n−1
X
= pk (t)Mk − pk+1 (t)Mk+1 = −pn (t)Mn = θn×n ,
k=1 k=0

which follows from Cayley-Hamilton theorem.

You might also like