Ma2aPracLectures 13-15

Ma2aPracLectures 13-15 - Lecture 13 Fundamental Matrices If...

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Unformatted text preview: Lecture 13 Fundamental Matrices If we have a linear system x = A x , with A an n n matrix with constant coecients a ij , 1 i, j n , a fundamental set , or basis , of solutions is given by x (1) , x (2) , . . . , x ( n ) such that every solution is of the form x = c 1 x (1) + c 2 x (2) + + c n x ( n ) , for suitable constants c 1 , c 2 , . . . , c n . Weve seen how to get a fundamental set of solutions ( x (1) , x (2) , . . . x ( n ) ) when A has n distinct eigenvalues 1 , 2 , 3 . . . n wth (column) eigenvectors v (1) , v (2) , . . . v ( n ) = 0 (i.e., A v ( i ) = j v ( j ) ): x ( j ) = v ( j ) e i t x ( j ) (0) = v ( j ) The associated fundamental matrix is given by ( t ) = ( x (1) x (2) . . . x ( n ) ) What is important here is not that A has distinct eigenvalues, but that there is a basis of n-space consisting of eigenvectors for A . In general there is no eigenbasis, and what we do know is that when the eigenvalues are all distinct, then there is definitely such a basis. Note : In general, there are many fundamental sets of solutions, and so ( t ) depends on the particular choice of x (1) , . . . x ( n ) . We have, for all j , d x ( j ) dt = A x ( j ) d dt = ( d x (1) dt , . . . , d x ( n ) dt ) n n matrix Thus satisfies the matrix differential equation d dt = A . Note : (0) = ( x (1) (0) . . . x ( n ) (0)) = ( v (1) , v (2) , . . . v ( n ) ). For example, one could have x (1) ( t ) = ( 2 e t e t ) and x (2) ( t ) = ( e 2 t e 2 t ) then v (1) = ( 2 1 ) and 1 v (2) = ( 1 1 ) , 1 = 1 and 2 = 2. In this case, = ( 2 e t e t e t e t ) , (0) = ( v (1) v (2) ) = ( 2 1 1 1 ) . Recall : We say that a fund matrix is in special form if (0) is the identity matrix I n . In this case, the eigenvectors are v (1) = 1 . . . = e 1 , v (2) = 1 . . . = e 2 , v ( n ) = . . . 1 = e n . When we have a special fundamental matrix, it is customary to denote it by ( t ) instead of ( t ). n = 3 : Suppose A is a diagonal matrix: 1 6 2 0 3 Eigenvalues : det( I 3 A ) = det 1 + 2 3 = ( 1)( + 2)( 3) = { 1 , 2 , 3 } For 1 = 1, A 1 = 1 2 = 2: A 1 = 2 = 2 1 3 = 3: A 1 = 3 1 = e t e 2 t e 3 t , (0) = I 3 2 For any n , the general solution x is a linear combination of fundamental ones. We can solve for the constants c 1 , c 2 , c 3 . . . c n , hence obtain a particular solution, if we are given the initial value x (0) = ( ....
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Ma2aPracLectures 13-15 - Lecture 13 Fundamental Matrices If...

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