Unformatted text preview: . Then the Jacobian of ρ a is the 2 × 3 matrix b J ( λ, x ) = £ s , (1 − λ ) I + λ J ( x ) ¤ , where J ( x ) is the matrix from Problem 1. (b) In order for the function to be transversal to zero, the matrix b J ( λ, x ) must be full rank (i.e., rank2) at every point λ ∈ [0 , 1), x, y ∈ ( −∞ , ∞ ). The matrix J ( x ) has two eigenvalues – call them α 1 and α 2 . The matrix K = (1 − λ ) I + λ J ( x ) has eigenvalues (1 − λ )+ λα i , so it is singular only if λ = 1 / (1 − α 1 ) or λ = 1 / (1 − α 2 ). Even if that happens, it is likely that the vector s will point in a di±erent direction, making the rank of J ( λ, x ) equal to 2....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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