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m2k_dfq_cstcof - Constant Coecient Linear Systems Denition...

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Constant Coefficient Linear Systems Definition A constant coefficient system of linear differential equa- tions takes the form Y prime ( t ) = A Y ( t ) + B ( t ) , where A is a constant n × n matrix. As usual, the system is homogeneous if B ( t ) 0, inhomogenous otherwise. Example 1 The system y prime 1 ( t ) y prime 2 ( t ) y prime 3 ( t ) = 2 1 0 1 3 1 0 1 2 y 1 ( t ) y 2 ( t ) y 3 ( t ) is a constant coefficient, first order, three dimensional, homogeneous system. When we have a constant coefficient scalar system z prime ( t ) = a z ( t ) the general solution takes the form z ( t, c ) = c e at , where c is an arbi- trary constant. Since the solutions of a vector system must be vector functions, it makes sense to try for a solution of the form Y ( t, λ, Φ) = Φ e λt , Φ R n , where λ is a constant (scalar) and Φ is a constant vector. Substituting this solution form into the system Y prime ( t ) = A Y ( t ) we have λ Φ e λt = A Φ e λt -→ ( λ I n - A ) Φ e λt = 0 . Since the exponential cannot vanish, we have ( λ I n - A ) Φ = 0 , i . e ., A Φ = λ Φ; 1
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thus A Φ is just a scalar multiple of Φ , λ being the scalar. Definition If λ is a number such that, for some non-zero n dimen- sional vector, Φ, ( λ I n - A ) Φ = 0 then λ is said to be an eigenvalue of the matrix A and Φ is an eigen- vector of A corresponding to λ . Proposition There exists a non-zero vector Φ such that ( λ I n - A ) Φ = 0 if and only if λ is such that det ( λ I n - A ) p ( λ ) = 0; here p ( λ ) is an n -th degree polynomial, p ( λ ) = λ n + a 1 λ n - 1 + · · · + a n - 1 λ + a n , which we call the characteristic polynomial of the matrix A . The fact that the determinant must be equal to zero for Φ negationslash = 0 to exist as shown is a familiar result from linear algebra. If we let the columns of A be A j and let the columns of the identity matrix be E j , j = 1 , 2 , ..., n, then p ( λ ) = det ( λ I n - A ) = det [ λ E 1 - A 1 , λ E 2 - A 2 , · · · , λ E n - A n ] = λ n det I n + · · · + ( - 1) n det A , using the linearity of the determinant with respect to the columns of the matrix in question and the form of p ( λ ) then follows. Example 2 For the 3 × 3 matrix in the system shown at the beginning of the lecture we have det ( λ I n - A ) = det λ - 2 - 1 0 - 1 λ - 3 - 1 0 - 1 λ - 2 = ( λ - 2)[( λ - 3)( λ - 2) - 1] - ( λ - 2) = λ 3 - 7 λ 2 + 14 λ - 8 2
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= ( λ - 1)( λ - 2)( λ - 4); the eigenvalues are 1 , 2 , and 4. It should be noted that eigenvectors are not quite unique; if Φ is an eigenvector of A corresponding to the eigenvalue λ and if c negationslash = 0 is a constant, then c Φ is also an eigenvector because it is a non-zero vector for which ( λ I n - A )( c Φ) = 0. Very commonly, in order to fix a particular eigenvector one chooses c = 1 bardbl Φ bardbl , in which case we obtain the eigenvector for which bardbl c Φ bardbl = vextenddouble vextenddouble vextenddouble vextenddouble 1 bardbl Φ bardbl Φ vextenddouble vextenddouble vextenddouble vextenddouble = 1 bardbl Φ bardbl bardbl Φ bardbl = 1. This process is called normalization and the resulting eigenvector is the normalized eigenvector.
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