V = (a1; a2) such that a1 and a2 are real and a2>0. is a vector space when we us
ID: 2938940 • Letter: V
Question
V = (a1; a2) such that a1 and a2 are real and a2>0.
is a vector space when we use the following operations of additionand scalar multiplication:
(a1, a2) + (b1, b2) = (a1 + b1, a2*b2); k(a1; a2) = (ka1;a^k2); k2 is a real.
I have already found my 0 vector and it should defined as (0,1) aslong as my calculations were correct. Which I am sure theyare. But I can't seem to show that it the below vectors areDependent. Any help would be greatly appreciated.
.
(a) Show that {(3, 4), (6, 16) } is linearly dependent.
Explanation / Answer
iff (3a + 6b, 4^a * 16^b) = (0, 1) iff a = -2 b.
So choose a = -2 and b = 1. Then
3a + 6b is -6 + 6 = 0 and 4^-2 * 16^1 = 16/16 = 1 which showsthat (3, 4) and (6, 16) are dependent because there is a linearcombination of the two equalling 0 for which the coefficients arenon-zero.
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