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Differential Equations Solutions 21

# Differential Equations Solutions 21 - 31 CHALLENGE 5.20(a...

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31 CHALLENGE 5.20. (a) Since U Σ V x = b , we have x = V Σ 1 U b . If we let c = U b , then α 1 = c 1 1 , and α 2 = c 2 2 . (b) Here is one way to look at it. This system is very ill-conditioned. The condition number is the ratio of the largest singular value to the smallest, so this must be large. In other words, σ 2 is quite small compared to σ 1 . For the perturbed problems, Ax ( i ) = b E ( i ) x ( i ) , so it is as if we solve the linear system with a slightly perturbed right-hand side. So, letting f ( i ) = U E ( i ) x ( i ) , the computed solution is x ( i ) = α ( i ) 1 v 1 + α ( i ) 2 v 2 , with α ( i ) 1 = ( c 1 + f ( i ) 1 ) 1 , α ( i ) 2 = ( c 2 + f ( i ) 2 ) 2 . From the figure, we know that f ( i ) must be small, so α ( i ) 1 α 1 . But because σ 2 is close to zero, α ( i ) 2 can be quite different from α 2 , so the solutions lie almost on a straight line in the direction v 2 . CHALLENGE 5.21. Some possibilities: Any right eigenvector u of A corresponding to a zero eigenvalue satisfies
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• Fall '11
• Dr.Robin
• Singular value decomposition, Orthogonal matrix, right singular vector, singular vector, largest singular value, eigenvalue satisﬁes Au

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