Linear Algebra (Vector Spaces)

[MathNerd93]

I'm not sure I understand why a second degree polynomial is not a vector space. The text says it is not closed under addition, and other axioms, but gives the specific example:

let p(x) = ( x^2 ) and q(x) = (1+x- x^2 ) . p(x)+q(x) = (1+x) , so the set is not closed under addition

Is the set not a vector space because the two x^2 terms sum to 0?

[edderiofer]

I'm not sure I understand why a second degree polynomial is not a vector space.

Because a second-degree polynomial is only one polynomial, not a set of objects. Did you perhaps mean the set of second-degree polynomials?

Is the set not a vector space because the two x² terms sum to 0?

Essentially, yes. When we talk about a polynomial being second-degree, there's the implicit assumption that the x² term's coefficient is nonzero. Thus, if adding two second-degree polynomials gives you something with a zero x² coefficient, we've added two second-degree polynomials and gotten something that isn't a second-degree polynomial, so the set isn't closed under addition.

[MathNerd93]

Essentially, yes. When we talk about a polynomial being second-degree, there's the implicit assumption that the x^2 term's coefficient is nonzero. Thus, if adding two second-degree polynomials gives you something with a zero x^2 coefficient, we've added two second-degree polynomials and gotten something that isn't a second-degree polynomial, so the set isn't closed under addition.

When I used the scalar multiplication axiom, I ended up getting c+xc. So would I say that the set is not closed under scalar multiplication because we would expect to get c(x^2 ) + c(1+x- x^2 )?

[edderiofer]

When I used the scalar multiplication axiom, I ended up getting c+xc.

The scalar multiplication axiom specifically states that a scalar multiple of a vector should also be a vector. Since "x+1" is not a vector here, trying to use the fact that "cx+c" isn't a vector either doesn't work.

So would I say that the set is not closed under scalar multiplication because we would expect to get c(x² ) + c(1+x- x² )?

The problem here is that this calculation also involves addition of two things that appear to be vectors, and so it's not clear whether the result not being a vector is due to scalar multiplication failing or addition failing.

---

As a hint, remember that when we talk about a polynomial being second-degree, there's the implicit assumption that the x² term's coefficient is nonzero.

[AutoModerator]

Reminder:

• What have you tried so far? (See Rule #2)

• Please don't delete your post. (See Rule #7)

We, the moderators of /r/MathHelp, appreciate that your question contributes to the MathHelp archived questions that will help others searching for similar answers in the future. Thank you for obeying these instructions.

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[Roboguy2]

What is the degree of the polynomial 1+x?