Skip to main content

Section 8.4 Application: Find eAt

It is very well known that
ex=1+x+x22!+x33!+
One can define
eA=I+A+A22!+A33!+
It is a fact that
I+A+A22!+A33!+
is convergent. In this class, we are able to compute eA if A is diagonalizable.
Example: Compute eA for A=(21215)
Solution:
  1. Find the characteristic polynomial det(AλI2)=λ2+3λ+2. The eigenvalues of A is 1,2.
  2. Find the corresponding eigenvectors. Thus we find a diagonal matrix D and an invertible matrix P such that
    P1AP=D or A=PDP1.
  3. Therefore,
    eA=ePDP1=I+PDP1+(PDP1)(PDP1)2!+=PP1+PDP1+PD2P12!+=P(I+D+D22!+)P1=P((1001)+(1002)+12!((1)200(2)2)+)P1=P(1+(1)+(1)22!+001+(2)+(2)22!+)P1=P(e100e2)P1=(4e3e212e212e1e1e24e23e)

Activity 8.4.1.

Supose that A is diagonalizable. How to compute eAt?
Solution.
Since A is diagonalizable, there exists an invertible matrix P and a diagonal matrix D=(λ1λn) such that P1AP=D. It is easy to see that
P1(At)P=P1(A(tIn))P=P1AP(tIn)=D(tIn)=tD.
That is,
eAt=P(eλ1eλn)P1
Exercise: compute eAt, where
A=(122252663)