chapter 4 answer - Chapter 4 General Vector Spaces Section...

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81 Chapter 4 General Vector Spaces Section 4.1 Exercise Set 4.1 1. (a) 12 34 1324 26 (,)( ,) (,) (, ) +=− + =−+ + = uv 3 u = 3( 1, 2) = (0, 6) (b) The sum of any two real numbers is a real number. The product of two real numbers is a real number and 0 is a real number. (c) Axioms 1 5 hold in V because they also hold in 2 . R (e) Let uu = u with 1 0. u Then 2 11 0 (, ) ( , ) . u == 3. The set is a vector space with the given operations. 5. The set is not a vector space. Axiom 5 fails to hold because of the restriction that x 0. Axiom 6 fails to hold for k < 0. 7. The set is not a vector space. Axiom 8 fails to hold because 222 () . km k m +≠ + 9. The set is a vector space with the given operations. 11. The set is a vector space with the given operations. 23. Hypothesis Add to both sides. Axiom 3 Axiom 5 Axiom 4 ( ) ( ) [( ) ] ) ] +=+ ++ −=++ − =++ = uw vw uw w vw w w uw w vw w u0 v0 25. (1) Axiom 7 (2) Axiom 4 (3) Axiom 5 (4) Axiom 1 (5) Axiom 3 (6) Axiom 5 (7) Axiom 4 True/False 4.1 (a) False; vectors are not restricted to being directed line segments. (b) False; vectors are not restricted to being n -tuples of real numbers. (c) True

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Chapter 4: General Vector Spaces SSM: Elementary Linear Algebra 82 (d) False; if a vector space V had exactly two elements, one of them would necessarily be the zero vector 0 . Call the other vector u . Then 0 + 0 = 0 , 0 + u = u , and u + 0 = u . Since u must exist and u 0 , then u = u and u + u = 0 . Consider the scalar 1 2 . Since 1 2 u would be an element of V , 1 2 = u0 or 1 2 . = uu If 1 2 , = then 11 1 22 . ⎛⎞ == = + =+= + ⎜⎟ ⎝⎠ u u u000 If 1 2 , = then 1 . + u u uuu0 Either way we get u = 0 which contradicts our assumption that we had exactly two elements. (e) False; the zero vector would be 0 = 0 + 0 x which does not have degree exactly 1. Section 4.2 Exercise Set 4.2 1. (a) This is a subspace of 3 . R (b) This is not a subspace of 3 , R since 121 2 2 2 ( ,,) ( ,,) ( , , ) aa a a += + which is not in the set. (c) This is a subspace of 3 . R (d) This is not a subspace of 3 , R since for k 1 () , ka c ka k c ++ ≠ + + so k ( a , b , c ) is not in the set. (e) This is a subspace of 3 . R 3. (a) This is a subspace of 3 . P (b) This is a subspace of 3 . P (c) This is not a subspace of 3 P since 23 01 2 3 ka ax ax ax ++ + is not in the set for all noninteger values of k . (d) This is a subspace of 3 . P 5. (a) This is a subspace of . R (b) This is not a subspace of , R since for k 1, k v is not in the set. (c) This is a subspace of . R (d) This is a subspace of . R 7. Consider 12 (, , ) . k k abc uv Thin 1 2 12 02 3 2 (, , ) ( , , ) . k k k k +−+ −= Equating components gives the system 2 2 kk b kk c = −+ = or A x = b where 01 21 , A ⎡⎤ ⎢⎥ = ⎣⎦ 1 2 , k k = x and . a b c ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ ⎣ ⎦ b The system can be solved simultaneously by reducing the matrix 2 3 0 0 4 0 2 5 5 0 , which reduces to 102400 012300 000010 . The results can be read from the reduced matrix. (a) (2, 2, 2) = 2 u + 2 v is a linear combination of u and v .
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chapter 4 answer - Chapter 4 General Vector Spaces Section...

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