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# s5 - Math 136 Section 3-4 Exercises Assignment 5 Solutions...

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Math 136 Assignment 5 Solutions Section 3-4 Exercises B3. a) S is not a subspace of R 4 since it does not contain the zero vector. b) We first observe T contains the zero vector and hence is a non-empty subset of R 4 . Let x, y T then ( 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 ) and ( x 1 + y 1 ) + 2( x 3 + y 3 ) = ( x 1 + 2 x 3 ) + ( y 1 + 2 y 3 ) = 0 and ( x 1 + y 1 ) - 3( x 4 + y 4 ) = ( x 1 - 3 x 4 ) + ( y 1 - 3 y 4 ) = 0 So T is closed under addition. For k R and x T We have k ( x 1 , x 2 , x 3 , x 4 ) = ( kx 1 , kx 2 , kx 3 , kx 4 ) and ( kx 1 ) + 2( kx 3 ) = k ( x 1 + 2 x 3 ) = 0 and ( kx 1 ) - 3( kx 4 ) = k ( x 1 - 3 x 4 ) = 0 , so T is closed under scalar multiplication. Therefore T is a subspace of R 4 . c) Two elements of U are x = (1 , 1 , 0 , 1) and y = (1 , 1 , 1 , 0) then x + y = (2 , 2 , 1 , 1) and ( x 1 + y 1 ) 2 = ( x 2 + y 2 ) 2 = 4, but ( x 3 + y 3 ) 2 + ( x 4 + y 4 ) 2 = 1 + 1 = 2. Thus x + y is not in U . Hence U is not closed under addition and therefore is not a subspace of R 4 . Note: Any example which shows that it is not closed under addition or is not closed under scalar multiplication is fine. B4. a) If ( - 4 , - 2 , 2 , - 6) can be expressed as a linear combination of vectors of V then there exists scalars x 1 , x 2 , and x 3 such that x 1 (1 , 1 , 0 , 1) + x 2 (2 , 0 , 0 , 2) + x 3 (0 , 2 , - 1 , 1) = ( - 4 , - 2 , 2 , - 6). Thus we are trying to solve the system with augmented matrix 1 2 0 - 4 1 0 2 - 2 0 0 - 1 2 1 2 1 - 6 . Row-reducing gives 1 0 0 2 0 1 0 - 3 0 0 1 - 2 0 0 0 0 . Hence the system is consistent and we have x 1 = 2, x 2 = - 3 and x 3 = - 2. i.e. 2(1 , 1 , 0 , 1) - 3(2 , 0 , 0 , 2) - 2(0 , 2 , - 1 , 1) = ( - 4 , - 2 , 2 , - 6) .

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2 b) If (6 , 0 , 0 , 3) can be expressed as a linear combination of vectors of V then there exists scalars x 1 , x 2 , and x 3 such that x 1 (1 , 1 , 0 , 1) + x 2 (2 , 0 , 0 , 2) + x 3 (0 , 2 , - 1 , 1) = (6 , 0 , 0 , 3). Thus we are trying to solve the system with augmented matrix 1 2 0 6 1 0 2 0 0 0 - 1 0 1 2 1 3 . Row-reducing gives 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 . Hence the system is not consistent so (6 , 0 , 0 , 3) is not in Sp( V ). B6 a) The nullspace of the linear mapping L with matrix A is the same as the solution space of the homogeneous system A x = 0 . Since there is no leading 1 in the third or fifth column of the matrix, x 3 and x 5 are arbitrary. let x 3 = s and x 5 = t . By back-substitution we get x 4 = - 4 t , x 2 = - 4 t and x 1 = t . Thus, the solution space is all vectors of the form ( x 1 , x 2 , x 3 , x 4 , x 5 ) = ( t, - 4 t, s, - 4 t, t ) = s (0 , 0 , 1 , 0 , 0) + t (1 , - 4 , 0 , - 4 , 1) so the nullspace of L is Sp { (0 , 0 , 1 , 0 , 0) , (1 , - 4 , 0 , - 4 , 1) } . b) The nullspace of the linear mapping L with matrix A is the same as the solution space of the homogeneous system A x = 0 . Since there is no leading 1 in the second, fourth or sixth column we have x 2 , x 4 , and x 6 are arbitrary. Let x 2 = r , x 4 = s and x 6 = t . Then by back substitution we have x 1 = 2 r + 4 s + 6 t , x 3 = 3 s + 6 t and x 5 = 5 t . Hence the solution space is all vectors of the form ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) = r (2 , 1 , 0 , 0 , 0 , 0) + s (4 , 0 , 3 , 1 , 0 , 0) + t (6 , 0 , 6 , 0 , 5 , 1) hence the nullspace of L is Sp { (2 , 1 , 0 , 0 , 0 , 0) , (4 , 0 , 3 , 1 , 0 , 0) , (6 , 0 , 6 , 0 , 5 , 1) } . B10. Since the range of L consists of all vectors that are multiples of (2 , 1), the columns of the matrix must be multiples of (2 , 1). Since the nullspace of L consists of all vectors that are multiplies of (2 , 1) we must have the dot product of this vector with any row equal to 0. Hence the rows of the matrix must be multiples of (1
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s5 - Math 136 Section 3-4 Exercises Assignment 5 Solutions...

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