Determine whether the following are subspaces of P4:
a) The set of polynomials in P4 of even degree
b) The set of all polynomials of degree 3
c) The set of all polynomials p(x) in P4 such that p(0) = 0
d) The set of all polynomials in P4 having at least one real root
The Attempt at a Solution
The book defines the vector space Pn as being all polynomials of degree n-1.
So I'm assuming that if R is a polynomial in P4,
[itex] R = r_3 x^3 + r_2 x^2 + r_1 x + r_0[/itex]
where [itex] r_n [/itex] is real.
For a set to be a subspace of P4 then, it must consist of polynomials of degree 3 or less, and according to the stipulations given in a), must be even degrees. Now I am not quite sure how to interpret this part. Does that mean the polynomial must contain elements of degree two or less? Or does this mean it must contain elements of only degree 2 and degree 0?
Regardless, when I test scalar multiplication and vector addition it seems to work fine in both cases, but the book says that only part c) is actually a subspace, and the rest are not.
[itex] \alpha R_2 = \alpha (r_2 x^2 + r_1 x + r_0) [/itex]
[itex] \alpha R_2 = \alpha r_2 x^2 + \alpha r_1 x + \alpha r_0 [/itex]
[itex] R_2 + Q_2 = (r_2 x^2 + r_1 x + r_0) + (q_2 x^2 + q_1 x + q_0) [/itex]
[itex] R_2 + Q_2 = ((r_2+q_2)x^2 + (r_1+q_1) x + (r_0+q_0))[/itex]
This resulting polynomial is still even degree, right? And even if I interpret the question the other way, meaning it does not contain r_1*x^1, this would still be true.
Is it because r2 can be anything? I could see that if r2 = 0, then the polynomial would be of odd degree in my initial interpretation (though not in my second interpretation) and would fail, but I'm not sure if that's what they are trying to get at.
Any advice would be appreciated!