# Is this set of polynomials a subspace of P2?

## Homework Statement

Let S denote the collection of all polynomials of the form p(t) = (2a - b)t^2 + 3(c - b)t + (a - c), where a,b,c are real numbers. Determine whether or not S is a subspace of P2.

## The Attempt at a Solution

Okay, so I know that in order for S to be a subspace, it must satisfy 3 conditions
1. closure of addition
2. closure of scalar multiplication
3. the set is not empty

I don't know how to show that S is a subspace of P2, my textbook does not give good examples. What should I do first?

## Answers and Replies

You must also show that it contains the 0 polynomial and that it is a subset of P2.
Simply use the definitions. Each of the elements in your subspace is determined by 3 numbers, so two example elements may be P(t)(a, b, c) and P(t)(d, e, f). Show that when you add these two, you get another polynomial of the form given.

HallsofIvy
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## Homework Statement

Let S denote the collection of all polynomials of the form p(t) = (2a - b)t^2 + 3(c - b)t + (a - c), where a,b,c are real numbers. Determine whether or not S is a subspace of P2.

## The Attempt at a Solution

Okay, so I know that in order for S to be a subspace, it must satisfy 3 conditions
1. closure of addition
2. closure of scalar multiplication
3. the set is not empty

I don't know how to show that S is a subspace of P2, my textbook does not give good examples. What should I do first?
Show that the 3 things you list are true!
1. Take a "vector" in S, u= (2a-b)t^2+ 3(c-b)t+ (a-c). Take another "vector" in S, v= (2a'- b')t^2+ 3(c'-b')t+ (a'-c'). I've used a', b', c' rather than a, b, c because they might be different numbers. What is u+ v? Is it in S? That is, is it of that same form?

2. Take a "vector" in S, u= (2a-b)t^2+ 3(c-b)t+ (a-c). Take a number, say, p. What is their scalar product? Is in in S? That's easy enough that I don't think it will hurt to show exactly what I mean: The scalar product is pu= p((2a-b)t^2+ 3(c-b)t+ (a-c))= p(2a-b)t^2+ 3p(c-b)t+ p(a-c)= (2pa- pb)t^2+ 3(pc- pb)t+ (pa- pc). That is again in S because it is also of the form "(2a-b)t^2+ 3(c-b)t+ (a-c)" with a replaced by pa, b replaced by pb and c replaced by pc. Therefore S is closed under scalar multiplication.

3. Choose 3 values for a, b, and c to find a specific "vector" that is in S.

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