I think the set has a zero vector and it\'s still closed under addition but if t
ID: 3408661 • Letter: I
Question
I think the set has a zero vector and it's still closed under addition but if the answer is 'C' which property does it fail? 4) Let H be the set of all polynomials of the form p(t) = a + bt2 where a and bare in and b > a. Determine whether H is a vector space. If it is not a vector space, determine which of the following properties it fails to satisfy. A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars A) H is not a vector space; not closed under multiplication by scalars and does not contain zero vector B) H is not a vector space; does not contain zero vector C) H is not a vector space; not closed under vector addition D) H is not a vector space; not closed under multiplication by scalarsExplanation / Answer
Let's evaluate the properties:
A. If the set contains a zero vector, then there has to exist the plynomial p(t) = 0 + 0t2. If this is true, then a=b=0, but in the problem it says that b>a, so a and b couldn't be zero at the same time, ergo, the set DOESN'T have a zero vector.
B. We take two vectors of H and sum:
(x + yt2) + (x' + y't2) = (x + x') + (y + y')t2
x, x', y and y' are in R, so, knowing that the sum of two real numbers is a real number, we can tell that the result of the sum is on the form a + bt2, and finally realize that H is closed under vector addition.
C. We take the vector x + yt2 which is in H, and the scalar z which is a real number, and make the multiplication:
z(x + yt2) = za + zyt2
The multiplication of real numbers results in a real number, so the vector za + zyt2 is in H, and then, it's closed under multiplication by scalars.
After this evaluation, we can say that the answer is C. The only property that fails is the existence of zero vector.