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Course by: Steven J. Cox. E-mail the author

# The Matrix Exponential via Eigenvalues and Eigenvectors

Module by: Steven J. Cox. E-mail the author

Summary: This module introduces how to compute the matrix exponential using eigenvalues and eigenvectors.

In this module we exploit the fact that the matrix exponential of a diagonal matrix is the diagonal matrix of element exponentials. In order to exploit it we need to recall that all matrices are almost diagonalizable. Let us begin with the clean case: if AA is nn-by-nn and has nn distinct eigenvalues, λjλj, and therefore nn linear eigenvectors, sjsj, then we note that A sj = λj sj   ,   j1n    j j 1 n A sj λj sj may be written

A=SΛS-1 A S Λ S
(1)
where S=( s1 s2 sn ) S s1 s2 sn is the full matrix of eigenvectors and Λ=diag λ1 λ2 λn Λ diag λ1 λ2 λn is the diagonal matrix of eigenvalues. One cool reason for writing AA as in Equation 1 is that A2=SΛS-1SΛS-1=SΛ2S-1 A 2 S Λ S S Λ S S Λ 2 S and, more generally Ak=SΛkS-1 A k S Λ k S If we now plug this into the definition in The Matrix Exponential as a Sum of Powers, we find eAt=SeΛtS-1 A t S Λ t S where eΛt Λ t is simply diage λ1 te λ2 te λn t diag λ1 t λ2 t λn t Let us exercise this on our standard suite of examples.

## Example 1

If A=( 10 02 ) A 1 0 0 2 then S=IΛ=A S I Λ A and so eAt=eΛt A t Λ t . This was too easy!

## Example 2

As a second example let us suppose A=( 01 -10 ) A 0 1 -1 0 and compute, in matlab,



>> [S, Lam] = eig(A)

S = 0.7071             0.7071
0 + 0.7071i        0 - 0.7071i

Lam = 0 + 1.0000i     0
0               0 - 1.0000i

>> Si = inv(S)

Si = 0.7071     0 - 0.7071i
0.7071     0 + 0.7071i

>> simple(S*diag(exp(diag(Lam)*t))*Si)

ans = [ cos(t),   sin(t)]
[-sin(t),   cos(t)]



## Example 3

If A=( 01 00 ) A 0 1 0 0 then matlab delivers



>> [S, Lam] = eig(A)

S = 1.0000   -1.0000
0         0.0000

Lam = 0    0
0    0



So zero is a double eigenvalue with but one eigenvector. Hence SS is not invertible and we can not invoke (Equation 1). The generalization of (Equation 1) is often called the Jordan Canonical Form or the Spectral Representation. The latter reads A=j=1h λj Pj + Dj A j 1 h λj Pj Dj where the λjλj are the distinct eigenvalues of AA while, in terms of the resolvent Rz=zIA-1 R z z I A , Pj =12πiRzdz Pj 1 2 z Cj R z is the associated eigen-projection and Dj =12πiRz(z λj )dz Dj 1 2 z Cj R z z λj is the associated eigen-nilpotent. In each case, CjCj is a small circle enclosing only λjλj.

Conversely we express the resolvent Rz=j=1h1z λj Pj +k=1 mj 11z λj k+1 D j k R z j 1 h 1 z λj Pj k 1 mj 1 1 z λj k 1 D j k where mj =dim Pj mj dim Pj with this preparation we recall Cauchy's integral formula for a smooth function ff fa=12πifzzadz f a 1 2 z C a f z z a where Ca C a is a curve enclosing the point aa. The natural matrix analog is fA=-12πifzRzdz f A -1 2 z C r f z R z where Cr C r encloses ALL of the eigenvalues of AA. For fz=ezt f z z t we find

eAt=j=1he λj t( Pj +k=1 mj 1tkk! D j k ) A t j 1 h λj t Pj k 1 mj 1 t k k D j k
(2)
with regard to our example we find, h=1 h 1 , λ 1 =0 λ 1 0 , P 1 =I P 1 I , m 1 =2 m 1 2 , D 1 =A D 1 A so eAt=I+tA A t I t A Let us consider a slightly bigger example, if A=( 110 010 002 ) A 1 1 0 0 1 0 0 0 2 then



>> R = inv(s*eye(3)-A)

R = [ 1/(s-1),   1/(s-1)^2,         0]
[       0,     1/(s-1),         0]
[       0,           0,   1/(s-2)]



and so λ 1 =1 λ 1 1 and λ 2 =2 λ 2 2 while P1 =( 100 010 000 ) P1 1 0 0 0 1 0 0 0 0 and so m 1 =2 m 1 2 D1 =( 010 000 000 ) D1 0 1 0 0 0 0 0 0 0 and P2 =( 000 000 001 ) P2 0 0 0 0 0 0 0 0 1 and m 2 =1 m 2 1 and D 2 =0 D 2 0 . Hence eAt=et( P1 +t D1 )+e2t P2 A t t P1 t D1 2 t P2 ( ettet0 0et0 00e2t ) t t t 0 0 t 0 0 0 2 t

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