Unformatted text preview: also showed, through extension of equalities, that the same conclusion holds even if A is not semisimple. (b) By (a), there is a unique symmetric solution to the equation. Let P = R ∞ e A T t Qe At dt . Then P > 0 since x T Px = R ∞ ( e At x ) T Q ( e At x ) dt and the integrand is > 0 whenever x 6 = 0 since e At is nonsingular and Q > 0. Further, P solves the equation since A T P + PA = Z ∞ d dt µ e A T t Qe At ¶ dt =Q. Exercise 4. (a) χ ( s ) = s 2 + (4 + 2 f 2 + f 1 ) s + (6 f 1 + 7 f 27). (b) χ ( s ) = s 2 + (7 k + 4) s + (27 k7). Exercise 5. k = ( a 1α 1 a 2α 2 a 3α 3 ) yields the desired characteristic equation....
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 Fall '10
 ClaireTomlin
 Linear Algebra, Matrices, Orthogonal matrix, 7K, 27K, 1 bu

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