Let P denote the set of all polynomials whose degree is exactly 2. Is P a vectro
ID: 2942198 • Letter: L
Question
Let P denote the set of all polynomials whose degree is exactly 2. Is P a vectro space? Justify your answer.This question is from section 4.2 #11 Book Differential Equations and Linear Algebra Third Edition (Goode, and Annin)
The answer is no and has been posted in this website under textbook help, however, there are three comments saying that there is a mistake but they are not very clear and all contradicting. Can you please explain this thoroughly, I assume it passes A1 but not A2 because if we multiply x^2(x^2) we get x^4 and since it has a different degree our answers dont look the same. Can you please answer and explain the question.
Thank you
Explanation / Answer
I'm not sure how you're axioms are labeled, but the set of polynomials of exactly degree 2 is not a vector space.
To show its not a vector space, you need to show that it fails at least one of the axioms. For instance, this set (let me call it P for now) is not closed under addition of polynomials. For instance, x2+5x+1 is in P, and so is -x2+2x-9. However, their sum is 7x-8, which is of degree 1 and not exactly degree 2. Thus, it isn't in the set P.
Another example: P is not closed under scalar multiplication because 0 times any polynomial of exactly degree 2 is 0, which has no degree. This also says that 0 isn't in P, showing that the axiom about the existence of 0 such v+0=0 for any v in V doesn't hold.
We can't use (x2)(x2)=x4 as an example because the main operations in our vector space in general are "vector addition" (in this particular example, polynomial addition) and "scalar multiplication", not "vector multiplication".
I hope this helps!
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