Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 17 a 2e i3 4e 2i3 50e i2 c 7e 3i2 49e 3i d

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Unformatted text preview: 1 D 2 2 cos nC1 .        2 1 1 2 5 14 16 A D produces u0 D , u1 D , u2 D , u3 D . This 1 2 0 1 4 13  1 is converging to the eigenvector direction with largest eigenvalue  D 3. Divide 1 uk by kuk k. 15 In the j th component of Ax 1 ; 1 sin Solutions to Exercises 94        121 12 15 1 14 1=2 17 A D gives u1 D , u2 D , u3 D ! u1 D . 1=2 312 31 94 27 13     1 cos  sin  cos .1 C sin2  / sin3  18 R D QT A D and A1 D RQ D . 0 sin2  sin3  cos  sin2  1 19 If A is orthogonal then Q D A and R D I . Therefore A1 D RQ D A again, and the QR method” doesn’t move from A. But shift A slightly and the method goes quickly to ƒ. 20 If A c I D QR then A1 D RQ C cI D Q in eigenvalues because A1 is similar to A. 1 .QR C cI /Q D Q 1 AQ. No change C bj q j C1 by q T to ﬁnd q T Aq j D aj (because the j j q ’s are orthonormal). The matrix form (multiplying by columns) is AQ D QT where T is tridiagonal. The entries down the diagonals of T are the a’s and b ’s. 21 Multiply Aq j D bj 1 q j 1 C aj q j 22 Theoretically the q ’s are orthonormal. In reality this important algorithm is not very stable. We must stop every few steps to reorthogonalize—or ﬁnd another more stable way to orthogonalize q ; Aq ; A2 q; : : : 1 AQ D QT AQ is also symmetric. A1 D RQ D R.QR/R D RAR has R and R 1 upper triangular, so A1 cannot have nonzeros on a lower diagonal than A. If A is tridiagonal and symmetric then (by using symmetry for the upper part of A1 / the matrix A1 D RAR 1 is also tridiagonal. P P 24 The proof of jj &lt; 1 when every absolute row sum &lt; 1 uses j aij xj j  jaij jjxi j &lt; jxi j. (Here xi is the largest component.) The application to the Gershgorin circle theorem (very useful) is printed after its statement in this problem. 23 If A is symmetric then A1 D Q 1 1 25 For A and K , the maximum row sums give all jj  1 and all jj  4. The circles j :5j  :5 and j :4j  :6 around diagonal entries of A give tighter bounds. The circle j 2j  2 p K contains the circle j 2j  1 and all three eigenvalues for p 2 C 2; 2, and 2 2. 26 With diagonal dominance ai i &gt; ri , the circles j ai i j  ri don’t include  D 0 (so A is invertible!). Notice that the 1; 2; 1 matrix is also invertible even though its diagonals are only weakly dominant. They equal the off-diagonal row sums, 2 D 2 except in the ﬁrst and last rows, and more care is needed to prove invertibility. 27 From the last line of code, q 2 is in the direction of v D Aq 1 h11 q 1 D Aq 1 .q T Aq 1 /q 1 . The dot product with q 1 is zero. This is Gram-Schmidt with Aq 1 as the 1 second input vector. 28 Note The ﬁve lines in Solutions to Selected Exercises prove two key properties of conjugate gradients—the residuals r k D b Ax k are orthogonal and the search directions are A-orthogonal .p T Ap i D 0/. Then each new guess x k C1 is the closest vector i to x among all combinations of b, Ab, Ak b. Ordinary iteration S x k C1 D T x k C b does not ﬁnd this best possible comb...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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