# Proving S is a subspace of P2

Let

S = { p $\in$ $P$$_{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}$ = 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 $\in$ 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.

Thanks

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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
p1=ax^2+bx+c
p2=dx^2+ex+f
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 $\cdot$ (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 $x\in\mathbb{R}$, $p(x)=0$. 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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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?

HallsofIvy
Homework Helper
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 $p_1(x)= a_1x^2+ b_1x+ c_1$ and $p_2(x)= a_2x^2+ b_2x+ c_2$ and $(p_1+ p_2(x))= (a_1+ a_2)x^2+ (b_1+ b_2)x+ (c_1+ c_2)$. To prove $(p_1+ p_2)(7)= 0$ you would have to "undo" that sum: $(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$$= (49a_1+ 7b_1+ c_1)+ (49a_2+ 7b_2+ c_2)= p_1(7)+ p_2(7)= 0+ 0= 0$.)

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).

Thanks once again HallsofIvy!