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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 coeﬃcients 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 . We’ve seen how to get a fundamental set of solutions ( x (1) , x (2) , . . . x ( n ) ) when A has n distinct eigenvalues λ 1 , λ 2 , λ 3 . . . λ n wth (column) eigenvectors v (1) , v (2) , . . . v ( n ) ̸ = 0 (i.e., A v ( i ) = λ j v ( j ) ): x ( j ) = v ( j ) e λ i t x ( j ) (0) = v ( j ) The associated fundamental matrix is given by Ψ( 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 nspace 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 Ψ( t ) depends on the particular choice of x (1) , . . . x ( n ) . We have, for all j , d x ( j ) dt = A x ( j ) ⇒ d Ψ dt = ( d x (1) dt , . . . , d x ( n ) dt ) ← n × n matrix Thus Ψ satisfies the matrix differential equation d Ψ dt = A Ψ . Note : Ψ(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 ) , λ 1 = 1 and λ 2 = 2. In this case, Ψ = ( 2 e t − e t − e t e t ) , Ψ(0) = ( v (1) v (2) ) = ( 2 − 1 − 1 1 ) . Recall : We say that a fund matrix Ψ is in special form if Ψ(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 Φ( t ) instead of Ψ( t ). n = 3 : Suppose A is a diagonal matrix: 1 6 − 2 0 3 Eigenvalues : det( λI 3 − A ) = det λ − 1 λ + 2 λ − 3 = ( λ − 1)( λ + 2)( λ − 3) = ⇒ λ ∈ { 1 , − 2 , 3 } For λ 1 = 1, A 1 = 1 λ 2 = − 2: A 1 = − 2 = − 2 1 λ 3 = 3: A 1 = 3 1 Ψ = e t e − 2 t e 3 t , Ψ(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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 Fall '08
 Makarov,N
 Linear Algebra, Matrices, Vector Space, Tn, dt

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