Question

Suppose that the set S in the hypothesis of Theorem GSP is not just linearly independent, but is also orthogonal. Prove that the set T created by the Gram-Schmidt procedure is equal to S . (Note that we are getting a stronger conclusion than ⟨ T ⟩ = ⟨ S ⟩ — the conclusion is that T = S .) In other words, it is pointless to apply the Gram-Schmidt procedure to a set that is already orthogonal.

Answer 1

Explanation:

Theorem Suppose v1, v2, . . . , vk are nonzero vectors that form an orthogonal set. Then v1, v2, . . . , vk are linearly independent.

Proof: Suppose t1v1 + t2v2 + · · · + tkvk = 0

for some t1,t2, . . . ,tk ∈ R.

Then for any index 1 ≤ i ≤ k we have

(t1v1 + t2v2 + · · · + tkvk , vi)= (0,vi)=0

⇒ t1(v1, vi) + t2(v2, vi) + · · · + tk(vk , vi)= 0

By orthogonality, ti(vi, vi) = 0 ⇒ ti=0.

Let V be a vector space with an inner product.

Suppose x1, x2, . . . , xn is a basis for V. Let

v1 = x1,

v2 = x2 −(x2, v1)/(v1, v1) v1

v3 = x3 −(x3, v1)/(v1, v1)v1 −(x3, v2)/(v2, v2)v2

Therefore,

vn = xn −(xn, v1)/(v1, v1) v1 − · · · −(xn, vn−1)/(vn−1, vn−1)vn−1.

Then v1, v2, . . . , vn is an orthogonal basis for V.

The orthogonalization of a basis as described above is called the Gram-Schmidt process.