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**Unformatted text preview: **+ x 2 y + xy-( y-a 2 ) ‚ . Then the Jacobian of ρ a is the 2 × 3 matrix 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 J ( λ, x ) must be full rank (i.e., rank-2) 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 diFerent direction, making the rank of J ( λ, x ) equal to 2. 2...

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- Fall '06
- oleary
- Linear Algebra, Rank, xy