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Differential Equations Solutions 19

Differential Equations Solutions 19 - 29 CHALLENGE 5.16...

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29 CHALLENGE 5.16. Form y = U b n 2 multiplications Form z = Σ 1 y n multiplications ( z i = y i i ) Form x = Vz n 2 multiplications Total: 2 n 2 + n multiplications CHALLENGE 5.17. (a) The columns of U corresponding to nonzero singular values form such a basis, since for any vector y , Ay = U Σ V y = n j =1 u j ( Σ V y ) j = σ j > 0 u j ( Σ V y ) j , so any vector Ay can be expressed as a linear combination of these columns of U . Conversely, any linear combination of these columns of U is in the range of A , so they form a basis for exactly the range. (b) Similar reasoning shows that the remaining columns of U form this basis. CHALLENGE 5.18.
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