Linear independent vector space

[realcomfy]

I have a quick question about vector spaces.

Consider the vector space of all polynomials of degree < 1. If the leading coefficient (the number that multiplies x^{N-1}) is 1, does the set still constitute a vector space?

I am thinking that it doesn't because the coefficient multiplying x^{N-1} is the same as the coefficient multiplying x^{0} = 1, and then it would not be linearly independent, or something like that, but I am not totally sure about this. Any clarification would be greatly appreciated.

 

[HallsofIvy]

I have a quick question about vector spaces.

Consider the vector space of all polynomials of degree < 1. If the leading coefficient (the number that multiplies x^{N-1}) is 1, does the set still constitute a vector space?

I am thinking that it doesn't because the coefficient multiplying x^{N-1} is the same as the coefficient multiplying x^{0} = 1, and then it would not be linearly independent, or something like that, but I am not totally sure about this. Any clarification would be greatly appreciated.

No, it is not a matter of being "linearly independent"- that is a property of a basis for a subspace, not the subspace itself. Polynomials in a subspace of polynomial can have the same coefficient for different powers. There is nothing wrong with that.

A subspace must have two properties:
a) It is closed under vector addition.
b) It is closed under scalar multiplication.

It should be easy to see that neither of those is satified by a set of polynomials with leading coefficient 1. If you add two such polynomials, you get a polynomial with leading coefficient 2, not 1. If you multiply such a polynomial by the number "a", you get a polynomial with leading coefficient "a", not 1.

By the way, did you really mean "the vector space of all polynomials of degree < 1"? That is the set of all constant functions and further requiring that "the leading coefficient is 1" reduces it to a single "vector"!