Let {v1, ..., vn} be a basis for a vector space V, and let L1 and L2 be two linear transformations mapping V into a vector space W. Show that if L1 (vi) = L2 (vi) for each i = 1, ..., n, then L1 = L2 [i. e., show that L1 (v) = L2 (v) for all v ∈ V].

Question

Keller

Mathematics

16 November, 18:13

Let {v1, ..., vn} be a basis for a vector space V, and let L1 and L2 be two linear transformations mapping V into a vector space W. Show that if L1 (vi) = L2 (vi) for each i = 1, ..., n, then L1 = L2 [i. e., show that L1 (v) = L2 (v) for all v ∈ V].

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Elliana Sutton · Answer 1

Lets take any arbitrary element v ∈ V, and lets prove that L1 (v) = L2 (v). Since {v1, ..., vn} is a basis of V, then v can be written as a unique linear combination of v1, ..., vn; in other words, there exists (unique) scalars a1, ..., an such that

a1 v1 + a2 v2 + ... + anvn = v

SInce both L1 and L2 are linears, we can take sum and product for scalars out of evaluation, thus

L1 (v) = L1 (a1v1 + a2v2 + ... + anvn) = a1 L1 (v1) + a2L1 (v2) + ... + anL1 (vn)

Since L1 (vi) = L2 (vi), we can replace L1 (v1) for L2 (vi) for each i in the expression above, obtaining the following expression still equal to L1 (v)

a1 L2 (v1) + a2L2 (v2) + ... + anL2 (vn) = L2 (a1v1 + a2v2 + ... + anvn) = L2 (v)

That shows that L1 (v) = L2 (v). Since v was arbitrary, then we have that L1 = L2.