Let v1 1 1 1 1T v2 1 2 1 2T and v3 0 1 0 2T and v v1 v2

Let v_1 = (1, 1, 1, 1)^T, v_2 = (1, 2, -1, 2)^T, and v_3 = (0, 1, 0, 2)^T and v = (v_1, v_2, v_3). Thus each v_i is in R^4 and v is a list of vectors. Use the Gram-Schmidt process to create an orthonormal list w = (w_1, w_2, W_3) with the same span as v.

Solution

Here v₁= [1 1 1 1]^T

v₂= [1 2 -1 2]^T

v₃= [0 1 0 2]^T . and v= {v₁, v₂, v₃}

Now if w= {w₁, w₂, w₃} is a orthonormal set of vectors, then assume w₁= v₁.

Let w₂ = a v₁+v₂ . Then (w₂, v₁)= 0 = a(v₁, v₁) + (v₂, v₁) i.e. a ||v₁||² + (v₂,v₁) = 0

=> a= - (v₂, v₁) / ||v₁||² .

=> a= - 4/4 = -1. So, w₂= v₂-v₁ = [0 1 -2 1]^T .

Now, let w₃= b.w₁ + c.w₂ + v₃ . Then (w₃, w₁) = 0 = b.|| w₁||² + c.(w₂, w₁) + (v₃, w₁)

=> b= - (v₃, w₁) / || w₁||² = - 3/4.

also (w₃, w₂) = 0 = b.(w₁, w₂)+ c.|| w₂||² + (v₃, w₂) => c = - (v₃, w₂) / || w₂||²

=> c = - 3/6 = -1/2

=> w₃ = (-3/4)w₁ + (-1/2)w₂ + v₃

=> w3 = [ -3/4 , -1/4 , 1/4 , 3/4]^T

Thus the orthonormal list w= { [1, 1, 1, 1]^T , [0, 1, -2, 1]^T , [-3/4, -1/4, 1/4, 3/4]^T}