By Peter J. Olver, Chehrzad Shakiban
Resolution guide for the ebook utilized Linear Algebra via Peter J. Olver and Chehrzad Shakiban
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One example is f (x) ≡ 0 and g(x) = x3 − x. 6. (a) f (x) = − 4 x + 3; (b) f (x) = − 2 x2 − x + 1. 7. ! 1 ex , which is a constant function. , and (a) 3 cos y ! − 5 x + 5 y − 5 ex − 5 x − y + ex + 1 . Multiplied by − 5 is (b) Their sum is − 5 x y − 5 cos y − 15 x y + cos y + 3 ! 0 . 8. This is the same as the space of functions F(R 2 , R 2 ). Explicitly: Commutativity of Addition: ! ! w1 (x, y) v1 (x, y) + w1 (x, y) w1 (x, y) v1 (x, y) + = = + w2 (x, y) v2 (x, y) + w2 (x, y) w2 (x, y) v2 (x, y) Associativity of Addition: !
Given any square matrix, write A = S + J where S = 12 A + AT is symmetric and “ ” J = 12 A − AT is skew-symmetric. This verifies the two conditions for complementary subspaces. 24(d). 28. (a) By induction, we can show that f (n) (x) = Pn 1 x ! e− 1/x = Qn (x) e− 1/x , xn where Pn (y) and Qn (x) = xn Pn (1/x) are certain polynomials of degree n. Thus, lim f (n) (x) = lim Qn (x) x→0 x→0 e− 1/x = Qn (0) y lim y n e− y = 0, →∞ xn because the exponential e− y goes to zero faster than any power of y goes to ∞.
C a = (c a1 , c a2 , c a3 , . . ). Explicity verification of the vector space properties is straightforward. An alternative, smarter strategy is to identify R∞ as the space of functions f : N → R where N = { 1, 2, 3, . . } is the set of natural numbers and we identify the function f with its sample vector f = (f (1), f (2), . . ). 11. (i) v + (−1)v = 1 v + (−1)v = 1 + (−1) v = 0 v = 0. (j) Let z = c 0. Then z + z = c (0 + 0) = c! 0 = z, and so, as in the proof of (h), z = 0. 1 1 1 (k) Suppose c = 0.