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Section 6.2 The Gram-Schmidt Process

2. [-1, 0, 1]*[1, 1, 1] = -1 + 0 + 1 = 0 so the generating set is orthogonal.
Then

b_W = (-1 + 0 + 3)/(1 + 0 + 1) [-1, 0, 1] + (1 + 2 + 3)/(1 + 1 + 1) [1, 1, 1]

    = 1[-1, 0, 1] + 2[ 1, 1, 1] = [1, 2, 3].


4. [1, -1, 1, 1]*[-1, 1, 1, 1] = 0,
[1, -1, 1, 1] * 1. 1. -1, 1] = 0, and
[-1, 1, 1, 1]*[1, 1, -1, 1] = 0 so the generating set is orthogonal.
Then

b_W = (1 - 4 + 1 + 2)/(1 + 1 + 1 + 1) [1, -1, 1, 1] +

      (1 - 4 + 1 + 2)/(1 + 1 + 1 + 1) [-1, 1, 1, 1] +

      (1 - 4 + 1 + 2)/(1 + 1 + 1 + 1) [1, 1, -1, 1]

    = 0[1, -1, 1, 1] + (3/2)[-1, 1, 1, 1] + (3/2)[1, 1, -1, 1]

    =[0, 3, 0, 3].


10. Applying the Gram-Schmidt process, we let v_1 = [1, 1, 1] and compute

[1, 0, 1] - (1 + 0 + 1)/(1 + 1 + 1) [1, 1, 1]

= [1, 0, 1] - (2/3)[1, 1, 1] = [1/3, -2/3, 1/3]

and take v_2 = [1, -2, 1].

[0, 1, 1] - (0 + 1 + 1)/(1 + 1 + 1) [1, 1, 1] - (0 - 2 + 1)/(1 + 4 + 1) [1,
-2, 1]

= [0, 1, 1] - (2/3)[1, 1, 1] - (1/6)[1, -2, 1] = [-1/2 0, 1/2]

and take v_3 = [-1, 0, 1].

An orthogonal basis is  {[1, 1, 1], [1, -2, 1], [-1, 0, 1]} so an
orthonormal basis is

{(1/3)^(1/2)[1, 1, 1], (1/6)^(1/2)[1, -2, 1], (1/2)^(1/2)[-1, 0, 1]}.


12. Again we apply the Gram-Schmidt process, letting v_1 = [1, -1, 1, 0, 0].
a_2 = [-1, 0, 0, 0, 1], so we get

a_2 - (a_2*v_1)/(v_1*v_1) v_1 = [-2/3, -1/3, 1/3, 0, 1],

and we take v_2 = [-2, -1, 1, 0, 3].

a_3 = [0, 0, 1, 0, 1], so we get

a_3 - [(a_3*v_1)/(v_1*v_1) v_1 + (a_3*v_2)/(v_2*v_2) v_2] = [1/5, 3/5, 2/5,
0, 1/5],

and we take v_3 = [1, 3, 2, 0, 1].

a_4 = [1, 0, 0, 1, 1], so we get

a_4 - [(a_4*v_1)/(v_1*v_1) v_1 + (a_4*v_2)/(v_2*v_2) v_2 +
(a_4*v_3)/(v_3*v_3) v_3]

= [2/3, 0, -2/3, 1, 2/3],

and we take v_4 = [2, 0, -2, 3, 2].

An orthogonal basis is
{[1, -1, 1, 0, 0], [-2, -1, 1, 0, 3], [1, 3, 2, 0, 1], [2, 0, -2, 3, 2]},

and an orthonormal basis is
{(1/3)^(1/2)[1, -1, 1, 0, 0], (1/15)^(1/2)[-2, -1, 1, 0, 3],
    (1/15)^(1/2)[1, 3, 2, 0, 1], (1/21)^(1/2)[2, 0, -2, 3, 2]}.



16. Let q_1 = (1/3)^(1/2)[1, -1, 1, 0, 0],
        q_2 = (1/15)^(1/2)[-2, -1, 1, 0, 3],
        q_3 = (1/15)^(1/2)[1, 3, 2, 0, 1],
        q_4 = (1/21)^(1/2)[2, 0, -2, 3, 2]
      and b = [-1, 0, 0, 1, -1].

Then b_W = (b*q_1)q_1 + (b*q_2)q_2 + (b*q_3)q_3 + (b*q_4)q_4

         = (-1/3)[1, -1, 1, 0, 0] - (1/15)[-2, -2, 1, 0, 3]
          - (2/15)[1, 3, 2, 0, 1] - (1/21)[2, 0, -2, 3, 2]

         = (-1/7)[3, 0, 4, 1, 3].


20. First we augment {a} = {[1, 1, 1]} to a basis for R^3.

[1 1 0 0]
|1 0 1 0| ~
[1 0 0 1]

[1 0  1 0]
|0 1  1 0|
[0 0 -1 1]

shows that {a, e_1, e_2} is a basis for R^3. Applying the Gram-Schmidt process,
let v_1 = a = [1, 1, 1], and a_2 = e_1, and we get

a_2 - (a_2*v_1)/(v_1*v_1) v_1 = [2/3, -1/3, -1/3]

so we take v_2 = [2, -1, -1].

a_3 = e_2, and we get

a_3 - [(a_3*v_1)/(v_1*v_1) v_1 + (a_3*v_2)/(v_2*v_2) v_2] = [0, 1/2, -1/2]

so we take v_3 = [0, 1, -1].

An orthogonal basis is {[1, 1, 1], [2, -1, -1], [0, 1, -1]},

and an orthonormal basis is
{(1/3)^(1/2)[1, 1, 1], (1/6)^(1/2)[2, -1, -1], (1/2)^(1/2)[0, 1, -1]}.



24. a) c_1 = [1, 2, -4]*[1/3, 2/3, -2/3] = 13/3
       c_2 = [1, 2, -4]*[2/3, 1/3, 2/3] = -4/3
       c_3 = [1, 2, -4]*[-2/3, 2/3, 1/3] = -2/3
       d_1 = [5, -3, 2]*[1/3, 2/3, -2/3] = -5/3
       d_2 = [5, -3, 2]*[2/3, 1/3, 2/3] = 11/3
       d_3 = [5, -3, 2]*[-2/3, 2/3, 1/3] = -14/3.

b) c*d = [13/3, -4/3, -2/3]*[-5/3, 11/3, -14/3] = -81/9 = -9

    [1, 2, -4]*[5, -3, 2] = -9.

  Thus the two dot products are equal.