Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?

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October 6th 2011, 07:53 PM
Pupil

Why 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?
October 6th 2011, 07:54 PM
Deveno

Re: 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.
October 6th 2011, 08:13 PM
Pupil

Re: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?

Quote:

Originally Posted by Deveno

there's nothing special about 2, we can define a similar space of polynomials of degree ≤ n, for any natural number n.

I know, but why is that $P_{n}$ is a subspace of $P_{n+1}$ ? My book is not very helpful at all with this, and would like to know the general reasoning behind this constraint.
October 6th 2011, 08:16 PM
roninpro

Re: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?

Quote:

Originally Posted by Pupil

I know, but why is that $P_{n}$ is a subspace of $P_{n+1}$ ? My book is not very helpful at all with this, and would like to know the general reasoning behind this constraint.

I think that we could better assist you if you could tell us what you understand about subspaces or how you define them.
October 6th 2011, 08:41 PM
Pupil

Re: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?

Quote:

Originally Posted by roninpro

I think that we could better assist you if you could tell us what you understand about subspaces or how you define them.

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.
October 6th 2011, 09:23 PM
roninpro

Re: 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 $P_2$ will again be a vector in $P_2$ (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 $P_2$ by a number will again be a vector in $P_2$ (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?
October 6th 2011, 09:43 PM
Pupil

Re: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?
Quote:
Originally Posted by roninpro

What "other rule" is confusing you?

To go back to the polynomial spaces example, let's try to consider the space $P_2$ (polynomials of degree 2 or less) within $P_3$ (polynomials of degree 3 or less). We can try to show that $P_2$ is a subspace of $P_3$ . We can use the conditions you provided. To rephrase, we must check:

Adding any two vectors in $P_2$ will again be a vector in $P_2$ (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 $P_2$ by a number will again be a vector in $P_2$ (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?
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 ?
October 6th 2011, 09:51 PM
roninpro

Re: 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 $P_2$ is a subspace of $P_3$ ? (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.)
October 7th 2011, 05:34 PM
Pupil

Re: Why must P_2 have a degree that ≤ 2 for it to be a subspace of P?

Quote:

Originally Posted by roninpro

Wait a second. We aren't showing that $P_2$ is a subspace of $P_3$ ? (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.)

That was my mistake, apologies. Ok, so $P_{2}$ is a subspace of $P_{3}$ if it holds for the two operations of vector addition and scalar multiplication, but what about $P_{3}$ being a subspace of $P_{2}$ ? Why would $P_{3}$ be considered a subset of $P_{2}$ but not a vector space? Is it due to the fact that adding two polynomials of degree two or less does not yield $P_{3}$ ?
October 7th 2011, 08:35 PM
Deveno

Re: 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.