Explanation:
Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that
a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.
Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.
Answer:
HCF=3
Step-by-step explanation:
Factors of 15t= 3*5*t
Factors of 6s= 2*3*s
HCF= Common factors
common factor = 3
HCF=3
Answer:
5
Step-by-step explanation:
*Just a highly recommended guess.*
Answer:
3.1 × 10^5
Step-by-step explanation:
8.37 ÷ 2.7= 3.1 when you divide you subtract the exponents
