Differential Equations Solutions 23

Differential Equations Solutions 23 - Chapter 6 Solutions:...

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Unformatted text preview: Chapter 6 Solutions: Case Study: Image Deblurring: I Can See Clearly Now (coauthored by James G. Nagy) Observe that if y is a p × 1 vector and z is a q × 1 vector CHALLENGE 6.1. then 2 y z p q 2 yi + = 2 i=1 2 zi = y 2 2 + z 2. 2 i=1 Therefore, g 0 − K αI 2 f 2 g − Kf −αf = 2 2 = g−Kf 2 2+ αf 2 2 = g−Kf 2 2 2 +α f 2 2 . CHALLENGE 6.2. First note that UT 0 Q≡ 0 VT is an orthogonal matrix since QT Q = I, and recall that the 2-norm of a vector is invariant under multiplication by an orthogonal matrix: Qz 2 = z 2 . Therefore, g 0 − K αI 2 f = 2 =Q 33 g − Kf −αf 2 2 g − UΣVT f −αf 2 2 ...
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