Section 2.4 Formulas for matrix-exponential for 3 by 3 matrices or smaller
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1 Page 1 Setion 2.4 Formulas for matrixexponential for 3 by 3 matries or smaller We will start with the simplest ase of a 2 by 2 matrix, and then extend the approah to 3 by 3 matries Our final objetive is to make it as easy as possible to ompute the exponential matrix for a matrix with two equal eigenvalues and a single one. Case I. Suppose that A is a 2 by 2 matrix with distint eigenvalues so that the harateristi polynomial is ² ³² ³² ³ with By definition of so ² ³ b an be omputed thusly ² ³b ² ³ ²³ where Notie that ² ³ Á ² ³Á Let ²³ so we see and similarly ²³ ² ³ ² ³ Reall where 7² ³² ³ ² ³ ² ³ ² ³ ²³ by definition. Hene b ( ²(³ ²(³ ² ³ ² ³ Using the def of Áwe get a formula for ( in terms of the eigenvalues. ( ( 0 ( 0 b (F ³
2 Example 1 Let (@ A Page 2 It follows from (F ³ ² ³ ² b ³² ³Á Á ( ( 0 (b0 ²³ b ² 0 (³b ²(b0³ Example 2 Let (@ A ² ³ ² ³² b ³ ( Thus (b0 b (0 9 (b0 ¹ ²³ ( 9 ² b ³²0 b ³¹ ² ³0 b ² ³²( ³ Case II. Suppose, i.e., ² ³² ³ Notie that here all the eigenvalues are idential. The best way to deal with this ase is by using CayleyHamilton theorem. However the purpose of the following rather ompliated approah is to shed light on the ase of mixed simple and multiple eigenvalues suh as the ase of ² ³² ³ ² ³ We will take the limit of (F ) as approahes First we let so that ² F ³ beomes ( ² b ³ ( 0 (² b ³0 b ² ³ (² b ³0 b ²( 0 ³µ ² ³ 0 b²( 0³ µ 0b²( 0³µ as In other words, if, then ( 0 b ²( 0³ µ (F ³ Case III. Consider a 3 by 3 matrix with distint eigenvalues, i.e.,
3 ² ³ ² ³² ³² ³ Page 3 Just as in ase I, we have ² ³ 7² ³² ³ ÁÁ ² for. ³ However ² ³ ² ³ More expliitly, ² ³² ³ 7 ² ³ ² ³ Á ² ³ ² ³² ³ ² ³² ³ 7 ² ³ ² ³ Á ² ³ ² ³² ³ ² ³² ³ 7 ² ³ ² ³ ² ³ ² ³² ³ By the relation obtained previously, 7 ²(³ ( ²(³ ² ³ b b ²( 0³²( 0³ ²( 0³²( 0³ ²( 0³²( 0³ (F³ ² ³² ³ ² ³² ³ ² ³² ³ Example 3 For (, find ( y ² ³ b b b b b ² ³ f f ² ³² ³ ² b ³² ³² ³ so we have Á and Substituting these eigenvalues into (F³Á we get ( ²(0³²(0³ ²(0³²(b0³b ²(b0³²(0³ For more expliit ( Á we ompute
4 ²(0³²(0³ Á y y y Page 4 ²(0³²(b0³ Á y y y ²(b0³²(0³ y y y Thus ( + b b b b z b } y b b b Case IV. A double and a single eigenvalues, e.g. ² ³² ³ ² ³. It is a limiting ase of (F ³ when approahes Let so that (F ³ beomes ( ²( 0³²( 00³ ²( 0³²( 0 0³ ² b ³ ²( 0³²( 0³ b b ² ³² ³ ² ³² ³ ² ³² b ³ Notie that the first and the last terms of the right side of the above equation beome ²( 0³ ² ³ ²( 0 0 ³ b 4 5 ²( 0 ³² b ³ µ ²( 0³ 0 ²( 0 ³ b ²( 0 ³² ² ³ b ² ³³µ ²( 0³ 0 b ²( 0 ³ ²( 0 ³² ² ³ b ² ³³µ ² ³ ²( 0³ ²( 0³ 0 b ²( 0³ µ as 0.
5 Together with the limiting value of the middle term involving, we have Page 5 or ( ²(0³ ²(0³ 0 b ²(0³ ³ ² ²( ³ 0 ³ µ b ² ³ (F ( ²( 0 ³ b ² ³²( 0 ³²( 0 ³µ b ²( 0 ³ ² ³ The last formula shows essentially two matrix multipliations Example 4 The last formula will be used to ompute ( for ( for whih ² ³ ² b ³ ² ³Á Á Á Á Á ( 0 ( 0 ²( 0 ³²( 0 ³ ²( 0 ³ Thus
6 ( b²b ³ b ²³ J K Page 6 x { x { x { ² ³ b b J K y y y b b b b y ²³ ²³ Remark Remember that we have relied on Cayley Hamiton theorem to deal with the ase of matries with idential eigenvalues. As a matter of fat, we have obtained ( 0 b ²( 0 ³ b ²( 0 ³ µ [ for suh a 3 by 3 matrix. Therefore we have developed formulas for (, in all possibilities, for matries of that size or smaller. Example 5 Let ( Compute the harateristi polynomial ( ) ² ³² ³ Let Á so that Á Á Using (F³Á we get
7 x { ²( 0³ Page 7 ²( 0³ ²( ²( 0³ 0³ Thus ²( 0³ ²( 0 ³ b ² ³²( 20 ³²( 0 ³µ b ²( 20 ³ ( ² 21³ b ² ³ µ b b b b y b Example 6 ( ² ³ ² ³² b ³ ² ³² b ³² ³ so we have Á Á Á three distint eigenvalues We use (F³ to get ( ²(b0³²(0³ ²(0³²(0³ ²(0³²(b0³ ²b³²³ ²³²³ ²³²b³ b b 9 ¹ b ²(b0³²(0³ ²( b0³ ²³²³ ²( 0 ³9 ² b ³²( b 0 ³ ¹ b ²( b 0 ³ ²³
8 Page 8 ²( 0 ³9 ² b ³ ²( b 0 ³ ²( 0 ³µ² ³³ b ²( b 0 ³ ²( 0 ³ ²( b 0 ³ b ²( 0 ³ ¹ b ²( b 0 ³ ²( 0³ ²(0³ b ²( b0³ b y This example demonstrates how we deal with omplex eigenvalues. However if the result omes out to be so simple, we should suspet that something is fishy here. Remember the purpose of omputing ( is to solve the system %W Z (%W For this (, the three equations are Z Z Z % &Á & %Á ' ' It is seen that the 3rd equation is not oupled to the 1st and 2nd equations. You should try to solve the system by method of elimination as an exerise. Exerises ( For Problems 1 to 3, ( ) ompute Á ²³ Z %W (%W with the initial ondition given by %W²³ find the solution of the initial value problem ( Á ( Á ( Á ZZZ ZZ Z Consider % b% % %. (i) Convert it to a first order system of three equations. (ii) Find the 3 by 3 matrix, A, of the system obtained in Part (i). (iii) Compute ( and hene the
9 general solution, x(t). (iv)then hek to see if the solution obtained in Part (iii) agrees with the solution found by the previous method disussed in Ch. 3. Page 9 Answers ( 1. ²0 b (( ³ b ²( ( b 0 ³ b ²( 0 ³ b b %²³ b ( 0 b²(0³b ²(0³ µ b b Á %²³ ( ²(b0³ ²(b0³²(0³ 0 b²(b0³µ x { x { x { b b Á y y y %²³ b 4. %²³ b b
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