speris1443
29.02.2020 •
Mathematics
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.
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Ответ:
Step-by-step 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.
Ответ: