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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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