Page 263 - 35Linear Algebra

14.4 Gram-Schmidt & Orthogonal Complements 263

Given a third vector w, we should first check that w does not lie in the
span{u, v}, i.e., check that u, v and w are linearly independent. If it does
not, we then can define

u w v ⊥ w
⊥
⊥
w := w − u − v .
u u v ⊥ v ⊥

⊥
⊥
We can check that u w and v ⊥ w are both zero:

⊥
u w v w
⊥
u w = u w − u − v ⊥
u u v ⊥ v ⊥
u w v ⊥ w
= u w − u u − u v ⊥
u u v ⊥ v ⊥
v ⊥ w
= u w − u w − u v ⊥ = 0
v ⊥ v ⊥

⊥
since u is orthogonal to v , and

⊥
u w v w
⊥
v ⊥ w = v ⊥ w − u − v ⊥
u u v ⊥ v ⊥
u w v ⊥ w
= v ⊥ w − v ⊥ u − v ⊥ v ⊥
u u v ⊥ v ⊥
u w
= v ⊥ w − v ⊥ u − v ⊥ w = 0
u u

⊥
⊥
⊥
because u is orthogonal to v . Since w is orthogonal to both u and v , we
⊥
⊥
have that {u, v , w } is an orthogonal basis for span{u, v, w}.
263