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Differential Equations Solutions 16

Differential Equations Solutions 16 - m n − 1 X k =1 k X...

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26 Chapter 5. Solutions: Matrix Factorizations We need to sum this from j = 1 to n , but we can neglect all but the highest order terms ( mn 3 and n 3 ), so only the cost of steps (5) and (6) are significant. For (5) we get n j =1 2( m j + 1)( n j + 1) mn 2 1 3 n 3 , since n j =1 j n 2 / 2 and n j =1 j 2 n 3 / 3. When m = n , this reduces to 2 n 3 / 3 + O ( n 2 ) multiplications. Determining the cost of (6) is left to the reader. For completeness, we include the operations counts for the other algorithms: Householder ( R only): for columns i = 1 : n , each entry of the submatrix of A is used once, and then we compute an outer product of the same size. n i =1 2( m i )( n i ) mn 2 n 3 / 3 . Givens ( R only): for columns j = 1 : n , for rows i = j + 1 : m , we count the cost of multiplying a matrix with 2 rows and ( n j ) columns by a Givens matrix. n j =1 m i = j +1 4( n j ) 2 mn 2 2 n 2 / 3 Gram-Schmidt : At each step k = 1 : n 1 there is one inner product and one
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Unformatted text preview: m . n − 1 X k =1 k X i =1 2 m ≈ mn 2 CHALLENGE 5.11. • Suppose z = ± b − A ˜ x ± ≤ ± b − Ax ± for all values of x . Then by multiplying this inequality by itself we see that z 2 = ± b − A ˜ x ± 2 ≤ ± b − Ax ± 2 , so ˜ x is also a minimizer of the square of the norm. • Since QQ ∗ = I , we see that || Q ∗ y || 2 2 = ( Q ∗ y ) ∗ ( Q ∗ y ) = y ∗ QQ ∗ y = y ∗ y = || y || 2 2 , Since norms are nonnegative quantities, take the square root and con-clude that || Q ∗ y || 2 = || y || 2 . • Suppose y 1 contains the Frst p components of the m-vector y . Then ± y ± 2 2 = m X j =1 | y j | 2 = p X j =1 | y j | 2 + m X j = p +1 | y j | 2 = ± y 1 ± 2 2 + ± y 2 ± 2 2 ....
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