Does this set of polynomials form a vector space?

[spaghetti3451]

Homework Statement

Consider the collection of all polynomials (with complex coefficients) of degree < N in x.

a) Does this set constitute a vector space? If so, suggest a convenient basis, and give the dimension of the space. If not, which of the defining properties does it lack?

b) What if we require that the polynomials be even functions?

c) What if we require that the leading coefficient (the number multiplying x^N-1) be 1?

d) What if we require that the polynomials have the value 0 at x = 1?

e) What if we require that the polynomials have the value 1 at x = 0?

Homework Equations

The Attempt at a Solution

a) I think it does constitute a vector space: the set is closed under addition and scalar multiplication. The zero vector is the polynomial with all coefficients = 0 and the inverse vector has coefficients with opposite sign.

The basis vectors are {xⁿ} for n = 0, 1, 2, ..., N-1.

The dimension of the space = N as there are N basis vectors.

b) if the polynomials are even functions, then each polynomial has only terms with even powers. Again, this set would satisfy the properties that define a vector space.

c) In this case, if we add two of them, the leading coefficient becomes 2, so the set is not a vector space.

d) Let's picture a graph of a polynomial. The only constraint on the trajectory of the curve is that the graph cuts the x-axis at x = 1. if we add two polynomials, the resultant polynomial still cuts the x-axis at x = 1. if we multiply the polynomial by a scalar, the resultant polynomial still cuts the x-axis at x = 1. The zero evctor is the zero polynomial and the inverse vector is the inverse polynomial. So, it's a vector space.

e) This time, if we add two polynomials, the resultant polynomial has y = 2 at x = 0, so clearly not a vector space.

I am hoping someone could provide feedback on the solution.

I woul be so happy if you do!

 

[LCKurtz]

It looks to me like you understand correctly which are vector spaces and which aren't. That's good. But if this is a problem you intend to hand in, you would need to justify your assertions with equations, even if it is real easy.

 

[spaghetti3451]

Thanks!