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Exercise 2.4.8

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Since $Q^{T} A = UD V^{T}$ , multiply both sides by $Q$ , we have $(QU) D V^{T} = Q Q^{T} A = Y Y^{+} A \approx A$ , the last step is due to Problem II.2.11 on page 135, where $Y = QR$ , so $Q Q^{T} = Y Y^{+}$ .

niuers

2020-03-20 00:00

Exercise 2.4.8

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