Can anyone please clarify why P_2 must have a degree ≤ 2 for it to be a subspace of P, the vector space of all polynomials?

- Oct 6th 2011, 07:53 PMPupilWhy must P_2 have a degree that ≤ 2 for it to be a subspace of P?
Can anyone please clarify why P_2 must have a degree ≤ 2 for it to be a subspace of P, the vector space of all polynomials?

- Oct 6th 2011, 07:54 PMDevenoRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
there's nothing special about 2, we can define a similar space of polynomials of degree ≤ n, for any natural number n.

- Oct 6th 2011, 08:13 PMPupilRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
- Oct 6th 2011, 08:16 PMroninproRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
- Oct 6th 2011, 08:41 PMPupilRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
Alright, I know that W is a subspace of V, a vector space, if W is also a vector space with respect to the operations in V. I also know that W is a subspace of V if and only if the following conditions hold:

1. if u and v are any vectors in W, then the vector addition of u and v is in W.

2. if c is any real number and u is any vector in W, then the scalar multiplication of c with u is in W.

W is a subspace of V if and only if it is closed under these two operations.

I also understand that every vector space has at least two subspaces, itself and the subspace 0 consisting only of the zero vector.

I understand all of this, just when they went into this other rule is where the confusion begins. - Oct 6th 2011, 09:23 PMroninproRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
What "other rule" is confusing you?

To go back to the polynomial spaces example, let's try to consider the space (polynomials of degree 2 or less) within (polynomials of degree 3 or less). We can try to show that is a subspace of . We can use the conditions you provided. To rephrase, we must check:

- Adding any two vectors in will again be a vector in (i.e. if I add two polynomials of degree two or less, the result is again a polynomial of degree two or less)
- Multiplying any vector in by a number will again be a vector in (i.e. if I multiply a polynomial of degree two or less by a number, the result is again a polynomial of degree two or less)

Do these two conditions hold? - Oct 6th 2011, 09:43 PMPupilRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
No, they do not. Hence, is not a subspace of . Thank you for explaining this. I finally understand.

One thing I don't understand, however, is the fact that although is not a subspace of , it is a subset of , no? If that is the case, how does one determine if a polynomial is a subset of ? - Oct 6th 2011, 09:51 PMroninproRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
Wait a second. We aren't showing that is a subspace of ? (You have written it in the reverse.) If the conditions fail, could you please give a counterexample? (That is, if you say that adding two polynomials of degree two or less does not necessarily give another polynomial of degree two or less, give an example.)

- Oct 7th 2011, 05:34 PMPupilRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
That was my mistake, apologies. Ok, so is a subspace of if it holds for the two operations of vector addition and scalar multiplication, but what about being a subspace of ? Why would be considered a subset of but not a vector space? Is it due to the fact that adding two polynomials of degree two or less does not yield ?

- Oct 7th 2011, 08:35 PMDevenoRe: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
P3 is NOT a subset of P2. x^3 is in P3, it is NOT in P2.