Consider the basis u1 1 1 1 u2 3 0 3 u3 3 3 0 Use the Gra

Consider the basis u_1 = [1 1 1] u_2 = [3 0 3] u_3 = [3 3 0] Use the Gram-Schmidt orthogonalization algorithm to transform u_1, u_2 and u_3 to an orthogonal basis v_1 v_2 and v_3. Convert v_1 v_2 and v_3 to an orthonormal basis w_1, w_2 and w_3. Let A be the matrix whose columns are w_1, w_2 and w_3. Check that A^T = A^-1. A matrix with this property is called an orthogonal matrix. The columns of an orthogonal matrix are an orthormal basis.

Solution

(a) v₁ = u₁ = (1,1,1)^T, v₂ = u₂ – proj_v1 (u₂) = u₂ –[(u₂.v₁)/(v₁.v₁)] v₁ = u₂ –[(3+0+3)/(1+1+1)] v₁ = (3,0,3)^T – 2(1,1,1)^T =(1,-2,1)^T and v₃=u₃ –proj_v1( u₃) - proj_v2( u₃) = u₃–[ (u₃ . v₁)/(v₁.v₁)] v₁ - [(u₃.v₂)/(v₂.v₂)] v₂ = u₃– [ (3+3+0)/(1+1+1)] v₁–[(3-6+0)/(1+4+1)]v₂ =(3,3,0)^T– 2(1,1,1)^T+1/2(1,-2,1)^T = ( 3/2 ,0,-3/2)^T.

(b) w₁ = v₁/||v₁|| = 1/3(1,1,1)^T = ( 1/3,1/3,1/3)^T, w₂ = v₂/||v₂||= 1/6(1,-2,1)^T = (1/6,-2/6, 1/6)^T and w₃ = v₃/||v₃||= 2/3 ( 3/2 ,0,-3/2)^T = ( 1/2, 0, -1/2)^T.

(c) We have A = [w₁ ,w₂ ,w₃] =

1/3

1/6

1/2

1/3

-2/6

0

1/3

1/6

-1/2

Then A^T =

1/3

1/3

1/3

1/6

-2/6

1/6

1/2

0

-1/2

Let B =

1/3

1 /6

1/2

1

0

0

1/3

-2/6

0

0

1

0

1/3

1/6

-1/2

0

0

1

The RREF of B is

1

0

0

1/3

1/3

1/3

0

1

0

1/6

-2/6

1/6

0

0

1

1/2

0

-1/2

Therefore A^-1 = A^T.

1/3	1/6	1/2
1/3	-2/6	0
1/3	1/6	-1/2