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:
your answer would be 2 and 1/5 :) hope this helps
Step-by-step explanation:
-Rayne
The rules for multiplying powers with the same base is you are basically doing the pemdas method Please Excuse My Dear Aunt Sally pemdas
Answer:
TBH IDRK
Step-by-step explanation:
You do what ever ask ur teacher not me im on here looking for answers myself lmbo
Answer: No. If two lines have the same slope they are parallel lines (or they are the same line), so they cannot intersect at one point. Because the triangles are congruent, the angles where the lines meet the x-axis are congruent. Because the lines form the same angle with the x-axis, they are parallel.
Step-by-step explanation:
