chapter5 - Chapter 5 Solutions to §5.1 1 v 1 =(6 2 v 2...

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Unformatted text preview: Chapter 5 Solutions to §5.1 1. v 1 = (6, 2), v 2 = (–3, 6), v 3 = (6, 2) + (–3, 6) = (3, 8). 3. Let x = (x 1 , x 2 , x 3 , x 4 ), y = (y 1 , y 2 , y 3 , y 4 ) be arbitrary vectors in R 4 . Then x + y = (x 1 , x 2 , x 3 , x 4 ) + (y 1 , y 2 , y 3 , y 4 ) = (x 1 + y 1 , x 2 + y 2 , x 3 + y 3 , x 4 + y 4 ) = (y 1 + x 1 , y 2 + x 2 , y 3 + x 3 , y 4 + x 4 ) = (y 1 , y 2 , y 3 , y 4 ) + (x 1 , x 2 , x 3 , x 4 ) = y + x . 5. Let x = (x 1 , x 2 , x 3 ) and y = (y 1 , y 2 , y 3 ) be arbitrary vectors in R 3 , and let r, s, t be arbitrary real numbers. Then: 1 x = 1(x 1 , x 2 , x 3 ) = (1x 1 , 1x 2 , 1x 3 ) = (x 1 , x 2 , x 3 ) = x . (st) x = (st)(x 1 , x 2 , x 3 ) = ((st)x 1 , (st)x 2 , (st)x 3 ) = (s(tx 1 ), s(tx 2 ), s(tx 3 )) = s(tx 1 , tx 2 , tx 3 ) = s(t x ). r( x + y ) = r(x 1 + y 1 , x 2 + y 2 , x 3 + y 3 ) = (r(x 1 + y 1 ), r(x 2 + y 2 ), r(x 3 + y 3 )) = (rx 1 + ry 1 , rx 2 + ry 2 , rx 3 + ry 3 ) = (rx 1 , rx 2 , rx 3 ) + (rx 1 , rx 2 , rx 3 ) = r x + r y . (s + t) x = (s + t)(x 1 , x 2 , x 3 ) = ((s + t)x 1 , (s + t)x 2 , (s + t)x 3 ) = (sx 1 + tx 1 , sx 2 + tx 2 , sx 3 + tx 3 ) = (sx 1 , sx 2 , sx 3 ) + (tx 1 , tx 2 , tx 3 ) = s x + t x . Solutions to §5.2 1. if x = p/q and y = r/s, where p, q, r, s are integers (q ≠ 0, s ≠ 0), then x + y = (ps + qr)/(qs), which is a rational number. Consequently the set of all rational numbers is closed under addition. The set is not closed under scalar multiplication, since if we multiply a rational number by and irrational number, the result is an irrational number. 136 Sec. 5.2 Definition of a Vector Space 137 3. V = {y : y´´ + 9y = 4x 2 } is not a vector space because it is not closed under vector addition. Let u, v ∈ V. Then u´´ + 9u = 4x 2 and v´´ + 9v = 4x 2 so it follows that (u + v)´´ + 9(u + v) = (u´´ + v´´) + 9(u + v) = u´´ + 9u + v´´ + 9v = 4x 2 + 4x 2 = 8x 2 ≠ x. Thus u + v ∈ / V. 5. V = { x ∈ R n : A x = , where A is a fixed matrix}. Addition and scalar multiplication is defined in R n . Let u , v ∈ V and k ∈ R . A1: Addition: A( u + v ) = A u + A v = + = so u + v ∈ V. A2: Scalar multiplication: A(k u ) = kA u = k = thus k u ∈ V. A3 and A4: Satisfied by properties of addition in R n . A5: = (0, 0, 0, …, 0) ∈ V, since A = . A6: If u ∈ V, then – u ∈ V since A(– u ) = –A u = – = . A7–A10: Satisfied by properties of matrix/vector algebra. Therefore, V is a real vector space. 7. (1) N is not closed under scalar multiplication, since multiplication of a positive integer by a real number does not, in general, result in a positive integer (2) There is no zero vector in N . (3) No element of N has an additive inverse in N . 9. Let A = a b c d e f be an arbitrary element in the set of all 2 × 3 matrices. Then we see that the zero vector is 0 2 × 3 = , since A + 0 2 × 3 = A. Further, A has additive inverse –A = –a –b –c –d –e –f since A + (–A) = 0...
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chapter5 - Chapter 5 Solutions to §5.1 1 v 1 =(6 2 v 2...

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