By Richard C. Penney

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8 that then {X1 , X2 , X3 } is independent and hence forms a basis. 43 (a) It is easily proven that the Xi are independent; hence span the nullspace. (b) [1, 1, 1, 1]t = 14 X1 + 14 X2 + 14 X3 , showing that [1, 1, 1, 1]t belongs to the nullspace. 45 Since A and B span, C is a linear combination of A and B; hence {A, B, C} is dependent. {B, C} need not be a basis of , for example C could be a multiple of B. 49 If = {0} then 0 spans . Hence assume ≠ {0}. 6 on page 115, can contain at most n linearly independent elements.

Hence X = 0. The independence follows. 17 Consider the equation xY1 + yY2 + xY3 = 0. Substituting the expressions for Yi into this equation yields x(X1 + 2X2 − X3 ) + y(2X1 + 2X2 − X3 ) + z(4X1 + 2X2 − X3 ) = 0 (2x + y + 2z)X1 + (3x − y + 13z)X2 + (5x + 3y + 3z)X3 = 0 There are an infinite number of solutions to the system x + 2y + 4z = 0 2x + 2y + 2z = 0 −x − y − z = 0 since the reduced form of its coefficient matrix is ⎡1 ⎢0 ⎢ ⎣0 0 −2⎤ 1 3⎥ ⎥ 0 0⎦ Hence the Yi are dependent. We do not need the independence of the Xi .

10 Since there is exactly one pivot entry in each non-zero row of the reduced form, the number of non-zero rows equals the number of pivot columns. Hence we expect the number of basis elements to equal the rank. 11 Suppose that x1 Y1 + x2 Y2 = 0 Then substituting the expressions for Yi yields x1 (3X1 − 2X2 ) + x2 (X1 + X2 ) = 0 (3x1 + x2 )X1 + (−2x1 + x2 )X2 = 0 It follows from the independence of the Xi that 3x1 + x2 = 0 −2x1 + x2 = 0 30 LINEAR INDEPENDENCE AND DIMENSION The augmented matrix for this system and its reduced form are respectively ] ] [ [ 1 0 0 3 1 0 0 1 0 −2 1 0 Hence x1 = x2 = 0, proving independence.