Applied Linear Algebra

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1 Applied Linear Algebra by Peter J. Olver and Chehrzad Shakiban Corrections to Instructor s Solution Manual..4d A =, x = u v, b = ; 0 w e A = 5, 5 f b =..4.5a Last updated: December 5, i a b and b 0; ii a = b 0, or a =, b = 0; iii a, b = 0..8.e 0,0,0 T ;..8a By induction, we can show that, for n and x > 0, f n x = Q n x x n where Q n x is a polynomial of degree n. Thus, Q lim x 0 fn x = lim n x + x 0 + x n e /x e /x, = Q n 0 lim y yn e y = 0 = lim x 0 fn x, because the exponential e y goes to zero faster than any power of y goes to..5.5b x =,,0 T, z = z 7, 7, T ;.5.d... while cokeru has basis 0,0,0, T. /5/ c 0 Peter J. Olver

2 .5.b Yes, the preceding example can put into row echelon form by the following elementary row operations of type #: 0 0 R R +R 0 R R R Indeed, Exercise.4.8 shows how to interchange any two rows, modulo multiplying one by an inessential minus sign, using only elementary row operations of type #. As a consequence, one can reduce any matrix to row echelon form without any row interchanges! The answer in the manual implicitly assumes that the row operations had to be done in the standard order. But this is not stated in the exercise as written..5.4 True. If kera = kerb R n, then both matrices have n columns, and so n ranka = dimkera = dimkerb = n rankb...6b... plane has length.....0c If v is any element of V, then we can write v = c v + +c n v n as a linear combination of the basis elements, and so, by bilinearity, x y,v = c x y,v + +c n x y,v n = c x,v y,v + +c n x,vn y,v n = 0. Since this holds for all v V, the result in part a implies x = y...b Missing square on v,w in formula: sin θ = cos θ = v w v,w v w = v w v w..4.ii andv Change null vectors to null directions..4.a L = A T T A T is....5.b 0 = Change all w s to v s. 4..4c When b, the minimum is Delete c. Just the label, not the formula coming afterwards a Change the interpolating polynomial to an interpolating polynomial. /5/ c 0 Peter J. Olver

3 4.4.5 ThesolutiongiveninthemanualisforthesquareS = { 0 x, 0 y }. When S = { x, y }, use the following: Solution:a z =, b z = x y, c z = One way to solve this is by direct computation. A more sophisticated approach is to apply the Cholesky factorization.70 to the inner product matrix: K = MM T. Then, v,w = v T Kw = v T ŵ where v = M T v, ŵ = M T w. Therefore, v,v form an orthonormal basis relative to v,w = v T Kw if and only if v = M T v, v = M T v, form an orthonormal basis for the dot product, and hence of the form determined in 0 Exercise 5... Using this we find: a M = 0 sinθ ±, for any 0 θ < π. b M = cosθ cosθ sinθ v = ±, for any 0 θ < π. cosθ 0, so v =, so v = cosθ sinθ p 0 x =, p x = x, p x = x, p 4 x = x 9 0 x. The solution given is for the interval [0,], not [,] iic 5.5.6iid , , v = cosθ+sinθ sinθ 5.6.0c The solution corresponds to the revised exercise for the system x +x +x = b, x +x = b, x +5x +7x = b, x +x +4x = b 4. For the given system, the cokernel basis is,,,0 T, and the compatibility condition is b +b +b = 0., /5/ c 0 Peter J. Olver

4 5.7.a,b,c To avoid any confusion, delete the superfluous last sample value in the first equation: a i f 0 =, f =, f =. ii e ix +e ix = cosx; b i f 0 =, f = 5, f = + 5, f = + 5, f 4 = 5; ii e ix e ix + e ix +e ix = cosx+cosx; c i f 0 = 6, f = +e πi/5 +e 4πi/5 = +.765i, f = +e πi/5 +e 4πi/5 = i, f = +e πi/5 +e 4πi/5 =.0777i, f 4 = +e πi/5 +e 4πi/5 =.765i; ii e ix ++e ix = +cosx+i sinx+cosx i sinx; d i f 0 = f = f = f 4 = f 5 = 0, f = 6; ii e ix +e ix e ix +e 4ix e 5ix = cosx+cosx cosx+cos4x cos5x+i sinx+sinx sinx+sin4x sin5x. 6..b The solution given in the manual corresponds to the revised exercise with incidence matrix Forthegivenmatrix,thesolutionis b u v u = f, u + v = g, u + u + v = f, u + v = g. 7.4.iib vt = c e t +c e t/ Set d = c in the written solution d Note: The solution is correct provided, for L:V V, one uses the same inner product on the domain and target copies of V. If different inner products are used, then the identity map is not self-adjoint, I I, and so, in this more general situation, L L. 8..a Interchange solutionsb andc., /5/ 4 c 0 Peter J. Olver

5 9.4.8 Change e t:a to e ta The solution given in the manual is for b =,,7 T. When b = 4,0,4 T, use the following: Solution: a x = = b x = 0, x = 0 4 c x k+ = d x = ; , x =, x = , with error e =.7500 x k + 0 ; , x = ; ; the error at the third iteration is e = ; the Gauss-Seidel approximation is more accurate e x k+ = 0 x k + ; f ρt J = 4 =.40, g ρt GS = =.8075, so Gauss Seidel converges about logρ GS /logρ J =.046 times as fast. h Approximately log /logρ GS 8.5 iterations. i Under Gauss Seidel, x 9 = , with error e 9 = Changeupperintegrationlimitintheformulafora = π y fzdz dy. π 0 0 b..f ϕx = δx 5 δx ; ϕxuxdx = u u for 5 a < < < b. a /5/ 5 c 0 Peter J. Olver

6 {, x >,..8d f x = 4δx++4δx +, x <, = 4δx++4δx + σx++σx, f x = 4δ x++4δ x δx++δx...a The symbol on the vertical axis should be n, not n...5 Replace σ y x by σ y x...7 In displayed equation, delete extra parenthesis at end of δ k ξ x twice; replace u by u in penultimate term: δ k ξ,u...9 Delete one repetition of On the other hand... a x y, 0 x y n, u n x = 4 nx + n xy 4 ny + y + x 4n, x y n, y x, y + n x. b Since u n x = Gx,y for all x y, we have lim n u n n x = Gx,y for all x y, while lim u n n y = lim y y n 4n = y y = Gy,y. Or one can appeal to continuity to infer this. This limit reflects the fact that the external forces converge to the delta function: lim f n n x = δx y. 0. c End of first line in final equation: I x + I z dα dx + I w dβ dx /5/ 6 c 0 Peter J. Olver

7 ..ci u x = x 5 +x, ii P[u] = iii P[u ] = 7 0 =.85, [ x u +x u ] dx, u = u = 0, iv P[x x] = 6 =.8, P[ sin πx] = div P [ 4 x+x+] = , P [ 0 xx+x+] = c Unique minimizer: u x = x+ +sinxcos log sincosx. The given solution is for the boundary conditions u = u = 0... ux = x 9 + logx 9 log.5. Change sign of last expression: = ω + e ω/ e ωx +e ω e ωx. ω +e ω/ sinhωy sinhωx, x < y,.5.7b For λ = ω ωsinhω < 0, Gx,y = sinhωx sinhωy, x > y; { ωsinhω xy, x < y, for λ = 0, Gx,y = yx, x > y; sinωy sinωx, x < y, for λ = ω n π ωsinω > 0, Gx,y = sinωx sinωy, x > y. ωsinω.5.9c i,ii,iii Replace b a by. Acknowledgments: We would particularly like to thank to David Hiebeler, University of Maine, for alerting us to many of these corrections. /5/ 7 c 0 Peter J. Olver

Applied Linear Algebra, First Edition

Applied Linear Algebra, First Edition by Peter J. Olver and Chehrzad Shakiban Corrections to First Printing (2006) Page xvii First bullet item The core: Add the following as a footnote: Last updated: July,