Linear Algebra Polynomial Vector Space

[Snippy]

Homework Statement

Use the subspace theorem to decide which of the following are real vector spaces with the usual operations.

a) The set of all real polynomials of any degree.
b) The set of real polynomials of degree \leq n
c) The set of real polynomails of degree exactly n.

Homework Equations

The Attempt at a Solution

I know how to do b) since the equation for the set of real polynomials of degree \leq n is:
P_n = {a₀ + a₁x + a₂x² + ... + a_nx² | a₀, a₁, ... , a_n \in R }

And I can prove that it is closed under addition and scalar multiplication.

But I am not sure what the difference between the equation for b) (at most n) and a) (any n) and c) (= n) is.

Also I know b) is a real vector space but I would've thought that meant c) was too since b) includes degree = n. But the answers say a) is b) is but c isn't.

 

[lanedance]

you need to check through all the axioms systematically... for c) is there an identity element? and is it closed?

 

[lanedance]

to see the difference between a) & b) consider the size of a basis set

 

[lanedance]

see here for the axioms, you should convince yourself they are either all satisfied, or see where the bust is...
http://mathworld.wolfram.com/VectorSpace.html

 

[Snippy]

My problem is I don't know what the difference between the equations for the 3 different problems is in order to check the axioms.

 

[lanedance]

ok so i would read it a polynomial of degree n, is any polynomial given by P_n = {a_0 + a_1x + a_2x^2 + ... + a_nx^2 | a_0, a_1, ... , a_n \in R}, when a_n is non-zero.

so
a) contains every Pm, for m = 0 to infinity
b) contains every Pm, for m = 0 to n
c) contains every Pm with m = n

though you also need to assume they contain 0... ie. in the n = 0 case, a_0 can be zero

 

[lanedance]

as some examples
x^3 + 1 is a polynomial of degree 3
2 is a polynomial of degree 0
and so on

 

[vela]

You have to consider n to be fixed. Say n=3 for example. Then x⁴ is an element of a) but is not an element of b) or c); x³ is an all of them; and x² is an element of a) and b) but not of c). Do you see why?