By Peter J. Olver, Cheri Shakiban

Recommendations handbook to utilized Linear Algebra. step by step for all difficulties.

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**Additional resources for Applied Linear Algebra - Instructor Solutions Manual**

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B) The space P of all polynomials, since every polynomial is a finite linear combination of monomials and vice versa. 21. (a) Linearly independent; (b) linearly dependent; (c) linearly dependent; (d) linearly independent; (e) linearly dependent; (f ) linearly dependent; (g) linearly dependent; (h) linearly independent; (i) linearly independent. 22. (a) The only solution to the homogeneous linear system 1 0 1 0 0 1 2 −2 1 C B C B B C 0C B −2 C B 3C C=0 C + c3 B B C + c2 B is c1 = c2 = c3 = 0. c1 B @ 1A @ −1 A @2A −1 1 1 (b) All but the second lie in the span.

3 y is x = y! = 0, the lines only ! x a 3b intersect at the origin. Moreover, every v = = + , where a = − 21 x+ 32 y, y a b b = 21 x − 12 y, can be written as a sum of vectors on each line. (c) A vector v = ( a, 2 a, 3 a )T in the line belongs to the plane if and only if a + 2 (2 a) + 3 (3 a)0 =114 a = 0 0, so a = 0 and the only 0 common element is1v = 0. Moreover, every 1 x x + 2 y + 3 z 1 B 1 B 13 x − 2 y − 3 z C C C v=B @ 2 (x + 2 y + 3 z) A + @ − 2 x + 10 y − 6 z A can be written as a sum @yA = 14 3 (x + 2 y + 3 z) 14 z −3x − 6y + 5z of a vector in the line and a vector in the plane.

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 · c v = (c v) = 0 = 0. (k) Suppose c = 0.