I think the set has a zero vector and its still closed under

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 scalars

Solution

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.