solution8 - M341 Homework Solution 8 Section 4.1 It is...

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M341 Homework Solution 8 Section 4.1: 4: It is obviously that the operations satisfy the closure properties. It remains to check eight properties. For x , y , and z R , we have x y = ( x 3 + y 3 ) 1 / 3 = ( y 3 + x 3 ) 1 / 3 = y x , and x ( y z ) = [ x 3 + ( y 3 + z 3 ) ] 1 / 3 = [ ( x 3 + y 3 ) + z 3 ] 1 / 3 = ( x y ) z , by the commutative law and the associative law of real numbers. Let 0 = 0 R , then for all x R we have x 0 = ( x 3 + 0 3 ) 1 / 3 = x = x . For any x R , let x = x R , then x ⊕ − x = [ x 3 + ( x ) 3 ] 1 / 3 = 0 1 / 3 = 0 = 0 . For x , y R and a, b R , we have a ( x y ) = 3 a ( x 3 + y 3 ) 1 / 3 = ( ax 3 + ay 3 ) 1 / 3 = [ ( 3 ax ) 3 + ( 3 ay ) 3 ] 1 / 3 = ( a x ) ( a y ) , ( a + b ) x = ( 3 a + b ) x = ( ax 3 + bx 3 ) 1 / 3 = ± ( 3 ax ) 3 + ( 3 bx ) 3 ² 1 / 3 = ( a x ) ( b x ) , and ( ab ) x = ( 3 ab ) x = ( 3 a )( 3 b x ) = a ( b x ) , 1 x = ( 3 1) x = x = x Therefore, R is the vector space under the operations. 6: Since the zero matrix O n is not in the set of nonsingular n × n matrices, the set does not have the identity element for addiction. Therefore the set is not a vector space. 7: The set R is not a vector space under the new operations since it doesn't satisfy the property (8), because for x = 1 R we have 1 x = 0 ̸ = 1 = x 9: The set R 2 is not a vector space under the new operations since it does not have the identity element for addiction. To see this, let x = [1 , 1] R 2 , then we have x y = [1 + y 1 , 0] ̸ = [1 , 1] = x for any choice of y = [ y 1 , y 2 ] R . 11: (a) Suppose B = 0 . For any X 1 and X 2 in V and a R , we have AX 1 = 0 and AX 2 = 0 , and it follows A ( X 1 + X 2 ) = AX 1 + AX 2 = 0 + 0 = 0 , A ( a X 1 ) = a ( AX 1 ) = a 0 = 0 Therefore the closure properties are satis ed. Conversely, if the closure properties are satis ed, let X ∈ V since the set is nonempty. Then we have AX = B and A (2 X ) = B by X and 2 X are all in V . It follows 1
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B = A (2 X ) = B = 2( AX ) = 2 B This implies B = 2 B B = 0 . (b) Properties (1), (2), (5), (6), (7), (8) have already been proven in Theorem 1.3 since V is a subset of M n 1 and the set equipped the usual operations as ones used in M n 1 . (c) It is easy to see that the identity element for addiction of
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solution8 - M341 Homework Solution 8 Section 4.1 It is...

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