# 6.7 - Math 2940 Solutions, Fall 2011 Section 6.7 10 21 110...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 6.7 10 21 110 110 1 1= . The charac6) With A = we have AAT = 12 011 011 01 teristic √ polynomial is λ2 −√ λ + 3 which has roots 3, 1. The corresponding unit eigenvectors 4 √ √ 2 are (1/ 2, 1/ 2) and (1/ , −1/ 2). Note σ1 = 3 and 2 = 1. 2 σ2 110 10 T 1 2 1 . The characteristic polynomial 1 1 1 1 0 = Also, A A = 011 011 01 3 2 is λ − 4λ + 3√ which √ roots√ 1 and 0.√ The corresponding unit eigenvectors are λ has 3, √ √ √ √ (1/ 6, 2/ 6, 1/ 6), (1/ 2, 0, −1/ 2), and (1/ 3, −1/ 3, 1/ 3). √ √ T √ √ √ √ 1/√6 1/ 2 1/√3 110 1/√2 1/√2 300 . So = 2/√6 √0 −1/√3 011 010 1/ 2 −1/ 2 1/ 3 1/ 6 −1/ 2 2 12) Since A = AT , A and AT A = A2 = AAT share the same eigenvectors. We have σi = λ2 i 30 so σ1 = 3 and σ2 = 2. So U = [u1 u2 ], Σ = and V = [u1 − u2 ]. We have a sign 02 change as σ2 = −λ2 . 13) We have R = U ΣV T so A = QR = (QU )ΣV T . The change is to U . ...
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## This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell University (Engineering School).

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