the exact methods to compute the matrix...

IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 12, Issue 1 Ver. IV (Jan. - Feb. 2016), PP 72-86 www.iosrjournals.org

DOI: 10.9790/5728-12147286 www.iosrjournals.org 72 | Page

The Exact Methods to Compute The Matrix Exponential

Mohammed Abdullah Salman1, V. C. Borkar

2

1(Department of Mathematics& statistics Yashwant Mahavidyalaya, Swami Ramamnand Teerth Marthwada

University, Nanded, India).

2(Department of Mathematics & Statistics, Yeshwant Mahavidyalaya, Swami Ramamnand Teerth Marthwada

University, Nanded, India).

Abstract: in this paper we present several kinds of methods that allow us to compute the exponential matrix

tAe exactly. These methods include calculating eigenvalues and Laplace transforms are well known, and are

mentioned here for completeness. Other method, not well known is mentioned in the literature, that don’t

including the calculation of eigenvectors, and which provide general formulas applicable to any matrix.

Keywords: Exponential matrix, functions of matrix, Lagrange-Sylvester interpolation, Putzer Spectral

formula, Laplace transform, Commuting Matrix, Non-commuting Matrix.

I. Introduction

The exponential matrix is a very useful tool on solving linear systems of first order. It provides a

formula for closed solutions, with the help of this can be analyzed controllability and ob servability of a linear

system [1]. There are several methods for calculating the matrix exponential, neither computationally efficient

[7,8,9,10]. However, from a theoretical point of view it is important to know properties of this matrix function.

Formulas involving the calculation of generalized Laplace t ransform and eigenvectors have been used in a large

amount of text books, and for this reason, in this work is to provide alternative methods, not well known,

friendly didactic. There are other methods [4] of at s interesting but not mentioned in the list of cases because of

its practicality in implementation. Eight cases or develop methods to calculate cases because of its practicality

in implementation. Eight cases or develop methods to calculate the matrix exponential. Provide examples of

how to apply the lesser-known methods in specific cases, and for the most known cases the respective

bibliography cited.

II. Definitions And Results

The exponential of an nn complex matrix A denoted bytA

e defined by

........... )!1(

)(......

!2

)(

!

)()(

1

0

2

n

AtAtAtI

k

Atet

n

k

k

At

To set the convergence of this series, we define firstly the frobenius norm of a matrix of size nm as follow

21

2

1 1

m

i

n

j

ijFaA

If ) (:, jA denotes the j-th column of A, and :) ,( iA the ith row, it is easy to see that is satisfy

21

1

2

2

21

2

21

:).()(:,

m

i

n

j

FiAjAA

We will use this standard for convenience, because in a finite dimension vector space all norms are equivalent.

An important property is to know how to use narrows the Frobenius norm of a matrix product. Given the

matrices nppm

BA

and then the product of them ABCij , with entries :) ,(:) ,( jBiAij . If A had

complex entries, we obtain conjugate ijC applied to row :) ,( iA . Recall the Cauchy-Schwarz inequality

22

):,(): ,( jBiACij

then we have :

22

F

2

21

2

21

2

2

2

21 1

2

1 1

2

j) B(:, ): ,()j , (:): ,(F

n

j

m

i

m

i

n

j

m

i

n

j

ijFBAiABiACAB

.

The Exact Methods To Compute The Matrix Exponential


𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑡𝑕𝑖𝑠 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑡𝑜 𝑎 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝐴 𝑖𝑠 𝑒𝑎𝑠𝑦 𝑡𝑜 𝑑𝑒𝑑𝑢𝑐𝑒 𝑡𝑕𝑒 𝑛𝑒𝑥𝑡 𝑓𝑜𝑟𝑚 :

,....3,2,1 allfor , nAAn

FF

n

Formally we must examine the convergence of the following limit:

lim𝑛→∞ 𝐴𝑘

𝑘!

𝑛

𝑘=0

It is sufficient to observe that satisfies:

FA

n

k

k

F

n

k

F

k

F

n

k

k

ek

A

k

A

k

A

000 !!!

and thus it demonstrated that 𝑒𝐴 already is well defined for any square matrix with constant entry. It is useful

to remember how the matrix exponential behaves under derivation:

.

..........!211

.............!

............!3

3

!2

2

..................!5!4!3!2!1

!)(

22

1322

55443322

0

AeAe

AttAIA

k

AktAttAA

AtAtAtAttAI

dt

d

k

At

dt

dt

tAtA

kk

k

kk

Using induction method and covenant condition )()( tt fo llows the formula:

o

ktAtAktA

k

k

)K(

Zk,AeeAedt

d)t( (2.1)

Note that the formula for the first derivative implies that the function o

tAxetx )( is solution of initial value

problem of the fo llowing first order system:

.)0( ,)(o

xxAxtx

Two results known of linear algebra that we used below are the fo llowing theorems.

Theorem (2.1) Schur Triangularization Theorem.

For any square matrixnn

A

, there is an unitary matrix U such that 1

UTUA is upper triangular.

In addition, the entries in a diagonal matrix T are the eigenvalues of A .

Theorem (2.2 ((Cayley Hamilton theorem)

Let A a square matrix and IA )( its characteristic polynomial then .0)( A

Now we will discuss several methods by details to calculate or compute and find the matrix

exponential.

III. Diagonalizable Matrix

Given a diagonal matrix nn ),.......,,(21 n

diagD It is to say that

)...........,.........,(21

k

n

kkkdiagD is true for all

Zk then we have

,,......,

!,.......,

!!

1

0 0

1

0

neediag

k

t

k

tdiagD

k

te

k

n

k k

k

k

k

k

k

k

tD



It is also a diagonal matrix. Now, in the case that A is a matrix diagonalizab le it is known that there exists an

invertible matrix P formed by the eigenvectors of A and a diagonal matrix D formed by the distinct

eigenvalues of A such that 1

PDPA . Now it is easy to verify the identity 1

PPDAkk

for all

Zk . Then we have:

.

!)(

!!

1

1

00

1

0

PPe

PDk

tPPPD

k

tA

k

te

tD

k

k

k

k

k

k

k

k

k

tA

Accordingly, it is trivial to find the exponential matrix of a d iagonalizable matrix, provided that previously we

find all the eigenvalues of A with corresponding eigenvectors. This case is well known.

IV. Not Diagonalizable Matrix

Suppose A is not diagonalizable matrix which it is not possible to find n linearly independent

eigenvectors of the matrix A , In this case can use the Jordan form of A . Suppose j is the Jordan form of A ,

with P the transition matrix. Then 1

PPeejA

Where

).....(),.....,,(1221112211 jjjdiagjjjdiagj

k

Then

)......( 2211 kkjjjJ

eeee

Thus, the problem is to find the matrix exponential of a Jordan block where the Jordan block has the form

kkkkMNJ )( and in general

kN as ones on the thk upper diagonal and is the null

matrix if nk the dimension of the matrix. By using the above expression we have

jjk

k

jk

kk

k

JN

J

k

kNI

kJ

ke k

000

)(

!

1)(

!

1

!

1

This can be written

)!1(.......

!2

12

n

NNNIee

n

j

Finally, we get the exponential matrix as fo llow:

1 PPee

jA

Then, 1

),........,( 1 PeePdiage m

JtJttA

This case is also well known. One way to demonstrate this formula is to consider the system of first order

Jxx with initial condition o

xx )0( . On the one hand, we know that the solution of this system is given

byo

Jtxetx

)( . Furthermore, this system is easy to solve, starting last equation, which is decoupled, then

every linear first-order equation is solved one by one, via the method of integrating factor.

V. Triangular Matrix:

Let S is an upper triangular matrix (lower triangular fo r a similar development is done) and write it as the sum

of a diagonal matrix with a n ilpotent matrix:

NDS

Recall that a matrix N is called a nilpotent if there is a positive integer r such that 0r

N . The smallest

positive integer, for which this equality holds, is called the index of n ilpotent of ownership matrix. Assuming

the known property on exponential matrices see [1]

, if, BAABeeeBABA

Now, we can use the above formula

to calculate:

.

)(s NtDtNDtteeee



We know how to calculate the matrix exponential of a d iagonal matrix, so we discuss how to get tN

e .

Sufficient to note that when N nilpotent, the series of this matrix becomes fin ite, since the number of a

quantity to be added to another is bounded by index of nilpotent of N . This at the same time is limited by the

degree of its min imal polynomial (remember that all nilpotent matrix has all zero) eigenvalues. We can

generalize this method to any parent. To do this, simply apply the theorem of Schur triangulation, 1

USUA

, where U is a unitary matrix and S is upper triangular. These yields:

1 UUee

StAt

and here we know how to proceed.

VI. Putzer’s Spectral Formula:

In [5], Putzer describes two methods to calculate tA

e .These are based on the fact that tA

e is a polynomial in

A whose coefficients t are scalar functions that can be found recursively by solving a simple system of linear

differential equations of the first order. We show only the second method, because it is easier to understand and

implement.

Theorem (6.1)

Given a matrix A of size nn , suppose we know all its eigenvalues n

,.......,,21

, not necessarily

distinct, but listed in a specific arbitrary order. Then it holds

1121)(................)()(

nno

tAPtrPtrPtre

where

121 P mstrixidentity where

1

k

n,...,,k,IA,IIP

k

j

jo

and )(........,),........(),(21

trtrtrn

are solutions of the differential system

0)0(r r

0)0(r r

1)0(r

n1-n

21222

1111

nnnrr

rr

rr

Proof:

put

,)()(

1

0

1

n

k

kkPtrt

and define 0)( tro

. Given kkkk

rrr 111

, compute

11

2

0

1

0

1

0

11

1

0

1

1

0

1)()()()(

nnnkk

n

k

nk

n

k

kk

n

k

kk

k

n

k

knk

n

k

kn

PrPrPrPr

PtrPtrtt

since the first term we can extract1nnn

Pr and the second term can be described as

1

0

11

1

1

1

0

,

n

k

kk

n

k

kk

n

k

kkPrPrPr

the right side of equality It simplifies to

. )(1

2

0

11

k

n

k

kknkrPP



Now, as is true thatkkk

PIAP )(11

, the expression in brackets is reduced to kn

PIA )( .

Furthermore it has

. - )(

, )( )( )(11

1

0

1

2

0

nnn

P

nnnkkn

n

k

kkn

n

k

rPIA

rPIArPIArPIA

n

But the Cayley-Hamilton theorem . 0n

P So, we obtained that )( - n

IAn

it implies that

. A Finally, as IPro )0()0(

1,it is fo llow

tAet )( by uniqueness of solutions.∎

As an application of this method, we find formulas for the exponential matrix of a matrix 22 in the fo rm

, ))(()(121IAtrItre

tA

with eigenvalues . ,21

According to the nature of the eigenvalues we have three cases to study.

a)- Real and distinct eigenvalues : We must solve the system of equations

1)0(r 1111

rr

0)0(r 21222

rrr

Solving the first equation, which is always decoupled, we get .)( 1

1

tetr

For the second, using the method

of integrating factor we obtain

. 1

- 1

)( 221

21

)(

21

2

tteetr

Consequently, we achieved the following formula

. 1

21

1

2

1 IAIee

Ieet

t

ttA

b) - Real and equal eigenvalues : In this case, to solve the system of equations we get t

etr 1)(1

and .)( 1

2

ttetr

So, we get the formula

. 1

11 IAteIeetttA

C)-Complex eigenvalues : In the case22

CA , with eigenvalues21

, , no problem in using the same

formula as in the case of real and distinct eigenvalues. But if A has real entries and its eigenvalues would be

complex conjugates, let say .0with , , 21

bibaiba In this case, to solve the system of

equations we get

b

bte

eetr

btibteetr

at

tt

att

)sin()(

,)sin()cos()(

21

2

1

21

1

where o

tAxetx )( is a solution for system ,Axx with in itial condition

oxx )0( , Ax

o and having

real entries, obviously we seek a real solution. So, the real part of the Spectral formula is to be considered

real solution, that is

, 121 o

tA

oxex)IA()t(rRI)t(rR)t(xR)t(x

and therefore it concludes that

. )sin(

)cos( aIAb

bteIbtee

atattA

Specific cases of Apostle:

In [2], Apostle shows how to obtain explicit formulas for the matrix exponential tA

e in the following

cases:

a)- all eigenvalues of A are equal,



b)- all eigenvalues of A are d istinct ,

c)- A has only two distinct eigenvalues, with one of mult iplicity algebraic one.

While these cases do not cover all possible alternatives for all of eigenvalues of a matrix, these

formulas will show by simplicity and because they help us to find all possible formulas for exponential matrices

of less than or equal to size 33 . It should be noted that the putzer's spectral formula also help us to deduce

these formulas, but the way obtained by Apostol is more forceful.

Theorem (7.1):

If A is a matrix nn with all its eigenvalues equal to then we have

. )(!

1

0

k

n

k

k

ttAIA

k

tee

Proof:

As the matrices I)-(A and ttI commute, we write

. )(!

)(

0

)( k

k

k

tIAtIttAIA

k

tIeeee

then the Cayley-Hamilton theorem implies 0 )( k

IA for nk , and so the theorem is proven.

Theorem (7.2):

If A is a matrix nn with n distinct eigenvalues n

,,,21

, then we have

, )(Lk

1

0

Aee

n

k

ttA k

where )( ALk

are the Lagrange interpolation coefficients given by

.,.......,3,2,1for )(

1

nkIA

AL

n

j jk

j

k

kj

Proof:

Although this theorem is a special case of the interpolation formula of Lagrange-Sylvester which gives us

directs proof. We define the following matrix function of scalar variab le as follow

. )()(

1

ALetF

k

n

k

tk

To prove tA

etF )( will show that F satisfies the differential equation )()( tAFtF with initial

condition F (0) = I. In fact, we observed that satisfy

).()()()(

1

k ALIAetFtAF

kk

n

k

t

By the Cayley-Hamilton theorem we have 0)(L)(k

AIAk

for each k , and so F satisfies

the differential equation. In addition to

n

k

kIALF

1

, )()0(

Finally, follow tA

etF )( uniqueness of solutions.

Theorem (7.3):

Let A be a matrix 3)(n , nn with two eigenvalues different and where has 1n

multip licity and has mult iplicity1 . Then it holds

. )()(!)()(

)(!

1

2

0

11

2

0

nk

n

k

k

n

t

n

t

k

n

k

k

ttAIA

k

teeIA

k

tee

Proof:

In scalar version, for fixed t , the expansion x

e of Taylor series is centered on t is



. )(k!0

k

k

t

xtx

ee

Now we evaluated tA and conveniently we partit ion this series

. )(i)!1-(n

)(k!

)(k!

)(k!

)(k!

)(k!

1

0

12

0

1

2

0

00

in

i

in

tk

n

k

k

t

k

nk

k

tk

n

k

k

t

k

k

k

tk

k

t

x

IAt

eIAt

e

IAt

eIAt

e

IAt

etItAe

e

Now we rewrite the second term of the last expression. As observe ,I)(IAIA

using the Cayley-Hamilton theorem, we have equality

,)IA)(()IA()IA()IA(01nn1n

that is . )IA)(()IA(1nn

By induction, it follows

,)IA()()IA(1n1iin

and so too

. I)- )(A()I-(A1

I)- (A)I-(A1

)IA()IA()IA(

1-nin

ininin

11

By rep lacing this relationship in the second sum we obtain

, 1

1

1

1

2

0

1

1

1

1

1

0

1

n

n

k

k

k

)(t

n

n

nk

k

k

n

n

i

i

in

)IA()(!k

te

)(

)IA()(!k

t

)(

)IA()()!in(

t

which completes the proof.

As we conclude the application of the matrix exponential formulas of any matrix A for size 3 x 3

according to the multip licity of its eigenvalues :

a - For eigenvalue with algebraic mult iplicity three have

.)IA(t)IA(tIeettA

22

2

1

b-For different eigenvalues ,,,321

have

.

2313

21

3212

31

3121

32 321

))((

)IA)(IA(e

))((

)IA)(IA(e

))((

)IA)(IA(ee

ttttA

c- For different eigenvalues121

with , has algebraic mult iplicity two have

. 2

1

12

2

12

12

1

112

1 )IA(te

)IA()(

ee)IA(tIee

ttt

ttA

Interpolation Lagrange-Sylvester and algorithm Gantmacher:



The next method, illustrated in Gantmacher [3], not only helps us to calculate the matrix exponential,

but also to evaluation of any analytic matrix function. First we mention the case where the eigenvalues of a

matrix are different, then the general case when there are multip licities.

Theorem (8.1):

If )A(f is a polynomial matrix nn

A

and if the eigenvalues of A are d ifferent, then )A(f it can be

decomposed as

,

1

n

i

ii)(z)(f)A(f

here n,.....,,ii 21

are the eigenvalues of A and )(zi

is a matrix nn is given by

.

1

n

k ki

k

i

ik

IA)(z

Proof:

By the Cayley-Hamilton )A(f can be reduced a polynomial of degree 1n , let say

. 2

2

1

1Ic............AcAc)A(f

o

n

n

n

n

This expression polynomial can be factored via interpolation of Lagrange which

j .

1 1

n

i

n

k

ki

ik

)IA(p)A(f

To calculatei

p , we mult iply it from the right both side of equality by the j -th eigenvector j

v corresponding

to j

.This procedure gives us

n

1k

n

1k1 1jkjk

, )-()-(jkjjjkjjj

n

i

n

k

kijvpvAvpv)IA(pv)A(f

ik

where the second equality is obtained by considering . 0ji

v)IA(

when the eigenvalues of A are different, we have jjj

v)(fv)A(f , and equality is easily deduced by

comparison

.

1

n

k

kj

j

j

jk

)(

)(fp

Therefore, all together it gives us

.

11

n

k ki

k

n

i

i

ik

IA)(f)A(f

When the eigenvalues are not distinct, algorithm of Gantmacher, see [3], can be used to expand the analytical

function )A(f as

,

1 1

1

jk

m

j

m

k

j

)k(

Z)!jk(

)(f)A(f

j

where m is the number of distinct eigenvalues , , jm is the multiplicity of the j-th eigenvalue.

)(fj

)k( is the derivative with respect to evaluated in the j-th eigenvalue and jk

Z are the

constituent matrices that are once found fixed for any analytic function )A(f . To illustrate the use of this

formula with an example .

Example 8.2: Let us find tA

e for the matrix



,

611

231

211

A

21 with multip licity 1

1m and 4

2 with multiplicity 2

2m . Then we have

.

11

222212111

2

2

1

2

1

1

1

1

1

1

2

1 1

1

Z)(fZ)(fZ)(f

Z)!k(

)(fZ

)!k(

)(f

Z)!jk(

)(f)A(f

k

k

)k(

k

k

)k(

jk

j

m

k

j

)k(j

Recall that this expansion is valid for any analytic function. In our situation, we are interested in the matrix

version of

t

e)(f . Here are the coefficients to be considered

, ,4

2

4

2

2

1

ttte)(fe)(fe)(f

To find the constituent matrices using known polynomial functions. As the matrix is of size 3 x 3, the Cayley -

Hamilton theorem states that tA

e must be have a polynomial function of 2

and , , AAI then will use the

following criteria.

If I)A(f ,the scalar version is I)(f . The coefficients are

. 0 , 1 ,1221

)(f)(f)(f We obtain the equation . 1211

ZZI

If A)A(f , the scalar version is )(f . The coefficients are

. 1 , 4 ,2221

)(f)(f)(f We obtain the equation . Z42222111

ZZA

If2

A)A(f , the scalar version is2

)(f . The coefficients are

. 8 , 16 ,4221

)(f)(f)(f We obtain the equation . Z8164222111

2 ZZA

Consequently, we must solve the formal matrix system

,

8164

142

011

2

22

21

11

A

A

I

Z

Z

Z

Then the solution is

2

2

2

2

22

12

11

2

134

4

123

4

124

21216

1812

1816

4

1

AAI

AAI

AAI

A

A

I

Z

Z

Z

Now, Put it all together we see that



tttttt

tttttt

tttttt

ttt

ttttA

ete))t(e(e))t(e(e

e)t(e)e)t(e())t(e(e

e)t(e))t(e(e))t(e(e

teee

ZteZeZee)A(f

242222

242422

242222

442

22

4

21

4

11

2

499144

19914

4

1

343234

17710

4

1

34399144

1956

4

1

42

7

2

7

42

1

2

5

42

7

2

3

04

9

4

9

34

3

4

7

34

9

4

5

14

9

4

9

34

1

4

7

34

9

4

9

It is the desired exponential matrix. It should be emphasizing that found once the constituent matrices; we can

find easily .,......... , )Asin()Acos( matrix functions of analytic functions.

Use of fundamental solutions of a linear differential equation with constant coefficients:

The following method avoids wonder if the matrix is diagonalizab le, and its prerequisites just know

Cayley Hamilton theorem and know how to solve linear homogeneous scalar differential equations of order n

with constant coefficients, like

01

2

2

1

1

xcxc.......xcxcx

o

n

n

n

n

)n(

Note that the method is not easy to implement if the roots of the polynomial equation (or characteristic

equation) associated with the above equation are uninteresting to get algebraically. The fo llowing theorems help

us to understand how this method works. The first theorem guarantees the existence and uniqueness of an

initial value problem for a matrix d ifferential equation, while the second provides a method for constructing the

solutions of matrix exponential based on initial value problems of scalar differential equations.

Theorem 9.1:

Let A be a nn constant matrix with characteristic polynomial

o

n

n

ncc.......c)AIdet()(p

1

1

1

Then tA

e)t( is the only solution of the matrix differential equation of order n given by

01

2

2

1

1

o

n

n

n

n

)n(cc.......cc (9.1)

with init ial condition

(0....., ,0 ,(0 , 01-n1-n20

A)A)(A)I)()(

(9.2)

Proof:

First we prove the uniqueness. Suppose )t()t(21

and are two solutions for (9.1) satisfying the initial

conditions given in (9.2). Define )t()t()t(21

- . Because of linearity, this function satisfies (9.1)

but with in itial condition

0(0....., ,0 (0 01)-(n0

))())()(

This means that each entry )t( satisfies the following initial value problem to be

ox....)(x)(x)(x

)t(xc)t(xc.......)t(xc)t(xc)t(x

)n(

o

n

n

n

n

)n(

1

1

2

2

1

1

000

0

where it is obvious that the only solution is )t(x 0 for all t , the trivial. Thus we have 0 )t( for all t ,

and obtain uniqueness. Now we prove the existence confirming that tA

e)t( satisfies the initial value

problem (9.1) - (9.2). Either the constant matrix with characteristic polynomial )(p disclosed in hypothesis .

Recall the fo rmula for k -th derivative of the exponential (see Equation (2.1))



...,.........,,,keA)t(tA)k()k(

3210 ,

We replaced on the right side of Equation (9.1) to obtain

01

2

2

1

1

tAtA

o

n

n

n

n

ne)A(pe)IcAc....AcAcA(

supported by the Cayley-Hamilton theorem. Finally, the formula of k -th derivative is deduced

(0....., ,0 ,(0 , 01-n1-n20

A)A)(A)I)()(

and )t( satisfies the initial conditions. This completes the proof.

Theorem 9.2:


o

n

n


1

1

1 Then we have

,A)t(x............A)t(xA)t(xI)t(xen

n

tA 12

321

where nk),t(xk

1 ,are the solutions of scalar differential equations n given by

01

1

xc................xcx

o

)n(

n

)n( (9.3)

and satisfying the initial conditions

1........,.................... 00

0010 00

0000 , 10

11

1

21

21

)n(

n

)n(

n

n

x,)(x

)(x,,)(x,)(x

)(x,,)(x)(x

(9.4)

Proof:

Define ,A)t(x........A)t(xA)t(xI)t(x)t(n

n

12

321

where )t(x

ksolutions of initial

value problems mentioned in the theorem. First we show that )t( satisfies equation (9.1). Indeed, we have

12

11

1

1

1

2

3

1

31

1

313

2

1

21

1

212

1

1

11

1

111

1

2

2

1

1

0

n

n

no

)(

n

)n(

nn

)n(

n

o

)()n(

n

)n(

o

)()n(

n

)n(

o

)()n(

n

)n(

o

n

n

n

n

)n(

oA.................OAOAI

A)xcxc.......xcx(

.....................................................................

A)xcxc.......xcx(

A)xcxc.......xcx(

I)xcxc.......xcx(

)t(c)t(c.......)t(c)t(c)t(

Now we show that satisfies Initial condition which g iven in (9.2):

1112

2

1

1

1

1

21

1

21

0............000

00 00

00 100

nn)n(

n

)n()n()n(

n

n

n

n

AA)(xA)(xI)(x)(

AA)(xA)(xI)(x)(

IA)(xA)(x)(x)(

Then, of uniqueness, is satisfied

.A)t(x............A)t(xA)t(xI)t(x)t(en

n

tA 12

321

for all t .Let us give this example to illustrate this method.

Example 9.3: We want to find the matrix exponential tA

e



,

611

231

211

A

For this we calculate that characteristic polynomial of A

.)AIdet()(p 32321023

We assume, that the Theorem (9.2) is satisfied

2

321A)t(xA)t(xI)t(xe

tA (9.5)

The characteristic polynomial produces the following scalar d ifferential equation with constant coefficients

0323210 xxxx

then the general solution is found based on the characteristic equation

042323210223 )m)(m(mmm

thus, the general solution takes the form ttt

teee)t(x4

3

4

2

2

1

To find )t(x1

we use the initial conditions .)(x,)(x,)(x 000010 by using this information we

obtain

.teee)t(xttt 442

1434

To find )t(x2

we use the initial conditions .)(x,)(x,)(x 001000 by using this informat ion we

obtain

.teee)t(xttt 442

2322

To find )t(x3

we use the initial conditions .)(x,)(x,)(x 100000 by using this information we

obtain

.teee)t(xttt 442

3

2

1

4

1

4

1

Finally, we replace these functions in formula (9.5) and got

.

ete))t(e(e))t(e(e

e)t(e)e)t(e())t(e(e

e)t(e))t(e(e))t(e(e

AtAteIt

tttttt

tttttt

tttttt

ttttttttttA

242222

242422

242222

2222222222

499144

19914

4

1

343234

17710

4

1

34399144

1956

4

1

)e2e-(1e4

1)32e(-2e)e43e-(4ee

It can be verified that the matrix A is not diagonalizable, but this is not relevant for calculat ions.

We can avoid some calculations in the above example? The following theorem sets the stage.

Theorem (9.4):


o

n

n


1

1

1 then we get the solution


n

tA 12

321

where



,

)t(

)t(

)t(

)t(x

)t(x

)t(x

)t(x

n

o

n

2

1

13

2

1

B (9.6)

where o

B is the evaluation here at t = 0, the matrix

,

)t()t()t(

)t()t()t(

)t()t()t(

)t(x

)t(x

)t(x

)t(x

)n(

nnn

)n(

)n(

o

n

1

1

222

1

111

13

2

1

B

provided that )t(),......,t(),t(Sn

21

is a fundamental set of solutions for the homogeneous linear

differential equation

01

1

xc................xcx

o

)n(

n

)n(

Proof:

Note that

01

1

1

o

n

n

ncc.......c)(p

is the characteristic equation of the differential equation

01

1

1

xcxc..............xcx

o

)n(

n

)n( (9.7)

Due to Theorem (9.2) we have


n

tA 12

321

where )t(xk

is solution of (9.7) with init ial conditions (9.4) for n,,.........,,k 321 .

we note the set )t(x),...,t(x),t(xn21

It is also a fundamental set of solutions of (9.3), since the Wronskian

))t(x),....,t(x),t(x(Wn21

is set takes the value 1 at t = 0. Furthermore, when using the known theorem of

uniqueness of solutions to the initial value problem (9.3) {(9.4), we have

)t(x)(....)t(x)()t(x)()t(x)()t(n

)n(

kkkkk0000

1

321

for each n,,.........,,k 321 , which we deduce

.

)t(x

)t(x

)t(x

B

)t(

)t(

)t(

n

o

n

2

1

2

1

As o

B is invert ible it is equal (9.6), which completes the proof.

Consider the matrix of Example (9.3) to appreciate the simplification to the calculations for the tabulation of tA

e . Recall that the characteristic polynomial of A is

.)AIdet()(p 32321023

which we associate the linear differential equation

0323210 xxxx



Such as characteristic equation is 0422 ))(( , we obtain

tttte,e,e

442

as a fundamental set of solutions. With these we form the matrix

,

21841

164

42

444

444

222

)t(e)t(ete

eee

eee

B

ttt

ttt

ttt

t

where it is calcu lated

.BBoo

211

1288

161216

4

1 inverse with

810

1641

421

1

then we get 2

321A)t(xA)t(xI)t(xe

tA , by Theorem 9.4 we have to be satisfied

.

2

1

4

1

4

1

322

434

442

442

442

4

4

2

1

3

2

1

ttt

ttt

ttt

t

t

t

o

teee

teee

teee

te

e

e

B

)t(x

)t(x

)t(x

By rep lacing these functions in the formula o f tA

e , the exponential matrix of Example 9.3 is recovered.

Laplace Transform: the Lap lace transform method is usual use in engineering to solve initial value problems

of scalar linear equations of order n with constant coefficients. This method is generalized to solve the problem

of in itial value .Axxo

x x(0), Applying Laplace transform we obtain

),(x)s(X)AsI()s(AX)(x)s(sX 0 that implies 0

such that we get . 01

)(x)AsI()s(X

After taking inverse Laplace transform we obtain

. 011

)(x)AsI(L)t(x

Thus, for uniqueness is achieved

. 11

)AsI(LetA

in applications of engineering, the exponential matrix is called the state transition matrix . For examples and

interesting properties of the exponential matrix, see [1].

VII. Conclusion

It illustrated several methods to calculate the exponential matrix of a square matrix. Most of them do

not use the calculation to eigenvalues (generalized) matrix, which has been a standard method of tackling the

problem in several kinds at the level o f init iation. While these methods can be applied to any matrix, all of them

are ineffective if dealing with large matrices or inaccurate entries (decimals truncated or numbers accompanied

with some error). This is where it is preferable to use the computer, but as the Work-Van Loan there is no

suitable method to numerical implementation.

References [1] P. J. Antsaklis & A. N. Michel; A Linear Systems Primer. Birkhauser, Boston (2007).

[2] T. M. Apostol; Some Explicit Formulas for the Exponential Matrix . tA

e the American Mathematical Monthly (1969), 76,

3:289{292. [3] F. R. Gantmacher; the Theory of Matrices, Volume I. Chelsea, New York (1959).

[4] S-H Hou & W-K. Pang; on the matrix exponential function. International Journal of Mathematical Education in Science and Technology (2006), 37, 1:65{70.

[5] E. J. Putzer; Avoiding the Jordan Canonical Form in the Discussion of Linear Systems with Constant Coefficients. The American

Mathematical Monthly (1966), 73, 1:2-7. [6] Nicholas J. Higham and Awad H. Al-Mohy, Computing Matrix Functions, Manchester Institute for Mathematical Sciences

2010,The University of Manchester, ISSN 1749-9097.pp1-47 [7] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press,1985, ISBN 0 -521-30586-2, xiii+561 pp.

[8] R. Bellman, Introduction to Matrix Analysis, 2nd ed. New York: McGraw- Hill,1960, reprinted by SIAM, Philadelphia,1995.



[9] Syed Muhammad Ghufran, The Computation of Matrix Functions in Particular, The Matrix Exponential, Master Thesis, University of Birmingham,England,2009,169 pp.

[10] Nathelie Smalls, The exponential of matrices, Master Thesis, University of Georgia, Georgia, 2007,49 pp.

the exact methods to compute the matrix...

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