Proving S is a subspace of P2

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In summary, Christoff provides a proof that S is a closed subspace of P_{2}(ℝ). He also provides a solution for the equation p(7)=0. However, he does not provide a proof that S contains the zero polynomial.
  • #1

Homework Statement


S = { p [itex]\in[/itex] [itex]P[/itex][itex]_{2}(ℝ) | p(7) = 0 }

Prove that S is a subspace of P[itex]_{2}(ℝ) (the vector space of all polynomials of degree at most 2)

The Attempt at a Solution

So essentially I have to prove that S is closed under addition, scalar multiplication and that the zero polynomial is in it.

1) Addition:

Let p1 = (x - 7)[itex]^{2}[/itex] = 0
Let p2 = (x - 7) = 0

Clearly p1(7) = 0 and p2(7) = 0

(p1 + p2)(7) = p1(7) + p2(7) = 0 + 0 = 0

Therefore S is closed under addition.

2) Scalar Multiplication:

λ [itex]\in[/itex] ℝ

(λp1)(7) = λp1(7) = λ(0) = 0

I am not sure whether the proof for scalar multiplication is correct or not.

3) Contains the zero polynomial

I do not know how to do this one.

Any help greatly appreciated.

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  • #2
You have an error in your proof for number 1. You seem to be choosing two PARTICULAR polynomials... You need to let p1 and p2 be arbitrary polynomials of degree at most 2 which satisfy p1(7)=p2(7)=0. eg... let
such that p1(7)=p2(7)=0 (the constants a,b,c,d,e,f can be ANYTHING such that this equality holds). Now show that their sum is again in S.

You proof for number 2 is correct, although you should consider using [itex]\cdot[/itex] (dot) to denote multiplication. When polynomials are involved, I typically consider a bracket to denote an evaluation, so in my mind the last line, λ(0) = 0, looks like you're evaluating some polynomial called lambda, at zero (when you're actually multiplying). This is just a matter of taste, and you can do with it what you will.

For number 3... What is the zero polynomial? It is the unique polynomial p(x) such that for all [itex]x\in\mathbb{R}[/itex], [itex] p(x)=0[/itex]. So, what is p(7)?

A discussion... Why you need to use arbitrary polynomials for your proof for number 1)... What about the polynomial p3(x)=x^2-4x-21 ? We have p3(7)=0, and it is of degree 2, so it is in S. But you haven't shown that the sum of IT with another polynomial in S is again in S. Your proof is incomplete. To prove it for all such polynomials, you must let p1 and p2 be arbitrary polynomials in S.
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  • #3
Hi christoff, thanks for your response.

Suppose for the addition proof I said:

Let two polynomials p1(5) = 0 and p2(5) = 0

Now, (p1 + p2)(5) = p1(5) + p2(5) = 0 + 0 = 0, would that be acceptable? I do not understand how to prove it using those coefficients you showed earlier. Although I will try now:

p1(x) = ax^2 + bx + c
p2(x) = dx^2 + ex + f

such that p1(7) = 0 and p2(7) = 0

Now (p1 + p2)(x) = (a + d)x^2 + (b + e)x + (c + f) = ax^2 + dx^2 bx + ex + c + f = p1(x) + p2(x)

∴ p1(7) + p2(7) = (49a + 14b + c) + (49d + 14e + f) = 0 + 0 = 0

...and therefore closed under addition? Doesn't seem right.

For the 3rd part, then p(7) = 0? I do not get how to show it contains the zero polynomial. Although if p(x) = 0 means the zero polynomial for any x, then p(7) = 0 already shows it contains it doesn't it?
  • #4
For any two functions, f and g, the definition of "f+ g" is that (f+g)(x)= f(x)+ g(x). In particular, if f(7)= 0 and g(7)= 0 then (f+g)(7)= f(7)+ g(7)= 0+ 0 = 0. You don't need to use cofficients.

(If you really want to, you would let [itex]p_1(x)= a_1x^2+ b_1x+ c_1[/itex] and [itex]p_2(x)= a_2x^2+ b_2x+ c_2[/itex] and [itex](p_1+ p_2(x))= (a_1+ a_2)x^2+ (b_1+ b_2)x+ (c_1+ c_2)[/itex]. To prove [itex](p_1+ p_2)(7)= 0[/itex] you would have to "undo" that sum: [itex](p_1+ p_2)(7)= (a_1+ a_2)(49)+ (b_1+ b_2)(7)+ (c_1+ c_2)= 49a_1+ 49a_2+ 7b_1+ 7b_2+ c_1+ c_2[/itex][itex]= (49a_1+ 7b_1+ c_1)+ (49a_2+ 7b_2+ c_2)= p_1(7)+ p_2(7)= 0+ 0= 0[/itex].)

For part 3, yes, if p is the 0 polynomial, then p(x)= 0 for all x and, in particular, p(7)= 0. That (and the fact that the 0 polynomial has degree less than 2) is all you need to show.

By the way, if you want to use "[ itex ]" you must end with "[ /itex ]" (without the spaces). And I think you will find it better to use them for entire formulas rather than just superscripts and subscripts (which can be done with the html tags [ sup ] [ /sup ] and [ sub ] [/sub] (again, without the spaces).
  • #5
Thanks once again HallsofIvy!

What is a subspace?

A subspace is a subset of a vector space that satisfies the same properties and operations as the larger vector space. This means that it contains the zero vector, is closed under addition and scalar multiplication, and is itself a vector space.

How do you prove that S is a subspace of P2?

To prove that S is a subspace of P2, we need to show that it satisfies the three requirements of a subspace: it contains the zero vector, is closed under addition, and is closed under scalar multiplication. This can be done by checking that all vectors in S can be written in the form ax^2 + bx + c, where a, b, and c are real numbers.

Why is it important to prove that S is a subspace of P2?

Proving that S is a subspace of P2 is important because it allows us to use the properties and operations of a vector space on S. This makes it easier to perform calculations and solve problems involving vectors in S.

What are some common mistakes when proving that S is a subspace of P2?

Some common mistakes include forgetting to check that the zero vector is included in S, assuming that all vectors in S are polynomials of degree 2 or less, and not showing that S is closed under addition and scalar multiplication.

Can S be a subspace of P2 if it does not contain the zero vector?

No, S must contain the zero vector in order to be a subspace of P2. This is because the zero vector is required for the closure under addition and scalar multiplication properties of a subspace to hold.