contestada

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].

Respuesta :

Answer:

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.

  • The calculation is as follows:

Lets considered any arbitrary element v ∈ V, and lets prove that

L1(v) = L2(v).

Since {v1 , ... , vn} depend upon  V, So 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 linear, now sum and product for scalars out of evaluation,

So,  

L1(v) = L1(a1v1 + a2v2 + .... + anvn) = a1 L1(v1) + a2L1(v2) + ... + anL1(vn)
Since L1(vi) = L2(vi), here we can replace L1(v1) for L2(vi) for each in the expression above, that obtained 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.

Learn more: https://brainly.com/question/26115859?referrer=searchResults