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Unformatted text preview: Lecture 16 For any m mmatrix M = ( m ij ) , 1 i, j m, the exponential of M is defined by exp( M ) = e M = n =0 M n n ! , when the right hand side converges, with M = I . Notation: n =0 ( a n b n c n d n ) = n =0 a n n =0 b n n =0 c n n =0 d n Why is this relevant for us? Reason : If we have x = A x , where x = x 1 . . . x m , x = d x dt = dx 1 dt . . . dx m dt , and A an m mmatrix with constant coecients, a basic set of solutions is given by the columns of the matrix ( t ) = e At Note : ( t ) = Ae At = A ( t ). A set of basic solutions are given by the columns of ( t ) = ( x (1) x (2) . . . x ( m ) ) , with x ( i ) ( t ) = x ( i ) 1 ( t ) x ( i ) 2 ( t ) . . . x ( i ) m ( t ) , i m. We have mostly looked so far at the case where A has distinct eigenvalues. A modification is necessary for repeated eigenvalues ! Now we will look at the exponential of some simple 2 2cases: 1 (i) A is diagonal : A = ( 1 2 ) A 2 = ( 2 1 2 1 ) , . . . , A n = ( n 1 n 2 ) So e A = A + A 1 1! + A 2 2! + + A n n ! + . . . = ( 1 0 0 1 ) + ( 1 2 ) + ( 2 2! 2 2 2! ) + + ( n 1 n ! n 2 n ! ) + . . . = ( 1 + 1 + 2 1 2! + + n n n ! 1 + 2 + 2 2 n ! + + n 2 n ! + . . . ) = ( e 1 e 2 ) Conclusion : exp ( 1 2 ) = ( e 1 e 2 ) Note : exp ( a b c d ) = ( e a e b e c e d ) ! (ii) A : upper triangular A = ( 1 0 1 ) A 2 = ( 1 2 1 ) A 3 = ( 1 3 1 ) , . . . , A n = ( 1 n 1 ) (by induction) Thus e A = n =0 A n n ! = n =0 ( 1 n ! n n ! 1 n ! ) = n =0 1 n ! n =0 n n ! n =0 1 n ! 2 n =0 1 n ! = e n =1 n n ! = n =1 1 ( n 1)! = k =0 1 k ! = e = e A = ( e e e ) Check : For A = ( 1 0 1 ) , e At = ( e t e t e t ) . (iii) A = ( 1 ) A 2 = ( 2 2 2 ) A 3 = ( 3 3 2 3 ) . . . A n = ( n n n 1 n ) Remembering that A = I , e A = n =0 n n ! n =0 n n 1 n ! n =0 n n ! Note that for any t , n =0 n n 1 t n n ! = n =1 n 1 t n ( n 1)! = t k =0 k t k k ! = te t Putting t = 1, e A = ( e e e ) Check : For A = ( 1 ) , e At = ( e t te t e t ) . 3 It appears hopeless to get a nice expression for e M for an arbitrary m m matrix M = ( m ij ), even for m = 2. We tackle this problem by appealing to similarity of matrices (also called conjugacy )....
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 Fall '08
 Makarov,N

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