(e) W = {x, 3x, 4x 3, 5x 4 x i R} in R 4 Math 33 Problem Set 4 Problems from.6 (pgs. 99- of text):,3,5,7,9,,7,9,,35,37,38. (Problems,3,4,7,9 in text). Determine whether the indicated subset is a subspace of the given R n. (a) W = {r, r r R} in R It suffices to show that if v and v are in W then so is any linear combination of v and v. Set v =r, r, v =r, r Then c v + c v = c r, r +c r, r = c r + c r, c r c r = (c r + c r ), (c r + c r ) W (b) W = {n, m n and n are integers} in R This subset is not closed under scalar multiplication for, W but, =, / W Since this subset is not closed under scalar multiplication it cannot be a subspace. (c) W = {x, y, z x, y, z R and z =3x +} in R 3 Consider two arbitrary vectors in W v =x,y, 3x +, v =x,y, 3x + we have v v =x x,y y, 3(x x )+ / W Since the difference of two vectors in W does not lie in W, W is not a subspace. (d) W = {x, y, z x, y, z R and z =,y =x} in R 3 Consider two arbitrary vectors in W v =x, x,, v =x, x, we have v v =x x, (x x ), / W
Consider two arbitrary vectors in R 4 x =x,x,x 3,x 4, x =x,x,x 3 4,x Then the vectors v =x, 3x, 4x 3, 5x 4, v =x, 3x, 4x 3, 4 5x will be in W. Wehave c v + c v = c x, 3c x, 4c x 3, 5c x 4 +c x, 3c x, 4c x 3, 5c x 4 = (c x + c x ), 3(c x + c x ), 4(c x 3 + c x 3), 5(c 4 x + c x 4) = x, 3x, 4x 3, 4 5x This vector belongs to W since x =c x + c x,c x + c x,c x 3 + c x 3,c 4 x + c x 4 R4 Since an arbitrary linear combinations of two vectors in W also lies in W, W is a subspace.. (Problem in text). Prove that the line y = mx is a subspace of R. (Hint: write the line as W = {x, mx x R}.) It suffices to show that an arbitrary linear combinations of two vectors in W also lies in W. Set v =x,mx, v =x,mx Then c v + c v = c x + c x,c mx + c mx Hence, W is a subspace. = (c x + c x ),m(c x + c x ) W 3. (Problems 7,9 and in text). Find a basis for the solution set of the following homogeneous linear systems. (a) 3x + x + x 3 = 6x +x +x 3 = 9x 3x 3x 3 = This linear system corresponds to the following augmented matrices 3 6 9 3 3 R 3 R R R 3 R 3 +3R 3 The latter augmented matrix corresponds to 3x + x + x 3 = = = Which is, effectively, one equation for three unknowns. Solving for x in terms of x and x 3 we obtain x = 3 (x + x 3 ) So any vector of the form 3 x 3 x3,x,x3 = x 3,, + x3 3,,
3 will be a solution. We conclude that e = 3,,, e = 3,, will be a basis for the solution space. (b) x + x + x 3 + x 4 = x 6x + x 3 = 3x 5x +x 3 + x 4 = 5x 4x +3x 3 +x 4 = This system corresponds to the following augmented matrix 6 3 5 5 4 3 Carrying out row reduction on this matrix yields R R R R 3 R 3 3 R R 4 R 4 5 R R 3 R 3 R R 4 R 4 R The last augmented matrix corresponds to which is equivalent to 3 3 3 3 x + x + x 3 + x 4 = 3 x + x 3 + x 4 = = = x = 3 (x 3 + x 4 ) x = (x + x 3 + x 4 )= ( 4 3 x 3 + 4 ) 3 x 4 = 7 3 (x 3 + x 4 ) Hence, any vector of the form 3 (x 3 + 7 x 4 ), 3 (x 3 + x 4 ),x 3,x 4 = x3 3 73,,, + 7 x 4 3, 3,, will be a solution. Thus, the vectors e = 7 3, 3,,, e = 7 3, 3,, provide a basis for the solution space of the original set of equations.
4 (c) x x +6x 3 + x 4 x 5 = 3x +x 3x 3 +x 4 +5x 5 = 4x +x x 3 +3x 4 x 5 = 3x x +4x 3 + x 4 8x 5 = x x +8x 3 +x 4 7x 5 = First we ll row-reduce the corresponding augmented matrix until it is in row-echelon form. 6 3 3 5 4 3 5 3 4 8 8 7 R R 3R R 3 R 3 4R R 4 R 4 3R R 5 R 5 R R R 5R 5 R 3 R 3 6R 5 R 4 R 4 R 5 R R 5 R 3 R 3 6R 5 6 5 8 6 4 5 4 5 6 33 9 4 5 6 4 5 9 33 R R 5 R 3 R 3 R 4 R 4 R 5 (/4)(R 5 R 3 ) 6 4 5 9 R R + R R 3 R 3 R 4 R 4 R 5 (/4)(R 5 R 3 ) 6 4 5 9 This last augmented matrix corresponds to the following linear system: x 5 = x 4 = x 3 = x 4 +9x 5 = x = 4x 3 +5x 5 = x = x 6x 3 x 4 + x 5 =
5 Thus, the only solution vector is =,,,,. One can think of the zero vector as the basis vector for the solution subspace, however, it s probably better to think of the solution space as the zero vector. 4. (Problems 35 and 37 in text). Solve the following linear systems and express the solution set in a form that illustrates Theorem.8. (a) () x x +3x 3 = 3 4x +4x x 4 = To solve the linear system we row reduce 3 4 4 From which we can infer R (/6)) (R R ) R R + R 3 3 6 6 x + x 3 x 4 = 7 6 3 7 6 or x x 3 6 x 4 = 7 6 x = x 3 + x 4 So a solution vector will have the form x 3 + x 4,x 3 + 6 x 4 + 7 6,x 3,x 4 Setting and and noting that h satisfies h = r,,, + s ( x x +3x3 = r + s ( 4x +4x x4 = 4 r + s x = x 3 + 6 x 4 + 7 6 = 7, 6,, + x3,,, + x4 6,,, p = 7, 6,,, 6,, = r + s,r+ s 6,r,s ) ( ) ( r + s +3r = r ( +3)+s 6 ) ( ) +4 r + s s = r ( 4+4)+s 6 We see that an arbitrary solution of (??) can be expressed as ) 6 ( 4 + 4 6 x = p + h with p a particular solution of (??) andh a solution of the corresponding homogeneous system. ) = =
6 (b) () x + x +3x 3 = 5 x x +x3 + x4 = 4x x +7x3 +x4 = 5 x x x 3 + x 4 = 5 The augmented matrix corresponding to this linear system is 3 4 7 This matrix can be row-reduced to the following form: which is equivalent to the following linear system or 3 5 5 5 5 x x 4 = x = x3 +x4 = 5 = x = 3x 4 + x = x3 = x4 5 Every solution vector can thus be written in the form x =3x4 +,, x4 5, x4 =,, 5, + x4 3,,, = p + h where p is a particular solution of (??) (the solution with x4 = ), and h =3r,, r, is a solution of the corresponding homogeneous system. 5. (Problem 38 in text). Mark each of the following statements True or False. a. A linear system with fewer equations than unknowns has an infinite number of solutions. False. If the system is inconsistent, it won t have any solutions. b. A consistent linear system with fewer equations than unknowns has an infinite number of solutions. True. c. If a square linear system Ax = b has a solution for every choice of column vector b, then the solution is unique for each choice of b. True. This follows from Theorems. and.6.
7 d. If a square system Ax = has only the trivial solution x =, then Ax = b has a unique solution for every column vector b with the appropriate number of components. True. This follows from Corollary and Theorem.6 e. If a linear system Ax = has only the trivial solution x =, thenax = b has a unique solution for every column vector b with the appropriate number of components. False. Consider what happens when the matrix A is of the form A = Then the corresponding system of equations x = x = = has a unique solution. Now consider a non-homogeneous linear system with the same matrix A; say Ax = b = The corresponding augmented matrix will be and so we ll have an inconsistent set of equations whenever b 3. f. The sum of two solution vectors of any linear system is also a solution vector of the system. False. Consider the following linear system: x + x = x + x3 = and the following two solution vectors v =,,, v =,, If we set v 3 = v + v =,, then, v 3 will not be a solution vector. g. The sum of two solution vectors of any homogeneous linear system is also a solution vector of the system. True. h. A scalar multiple of a solution vector of any homogeneous linear system is also a solution vector of the system. True. The solution space of a homogeneous linear system is a subspace: therefore it s closed under scalar multiplication. i. Every line in R is a subspace of R generated by a single vector. b b b 3 b b b3
8 False. The line { x R x =(, ) + (, )t ; t R } is not a subspace. j. Every line in R through the origin in R is a subspace of R generated by a single vector. True.