Show that S is the subspace of P3

Question

I'm revising for my linear exam and came across this question:

Consider the vector space P3 of all polynomials of degree <= 3, and consider the subset S of P3 given by

$$\text{ {S = a + b$x^2$ + c$x^3$ | a - b = 0} }$$

Show that S is a subspace of P3

So I started by checking the first axiom (closed under addition) to see if S is a subspace of P3: Assume

polynomial 1 = $a_1 + b_1 x^2 + c_1 x^3$

polynomial 2 = $a_2 + b_2 x^2 + c_2 x^3$

p1+p2 = $a_1 + b_1 x^2 + c_1 x^3$ + $a_2 + b_2 x^2 + c_2 x^3$

= $ (a_1 + a_2) + (b_1 + b_2) x^2 + (c_1 + c_2) x^3$

But since a - b = 0, then a = b which means

= $ (a_1 + a_2) + (a_1 + a_2) x^2 + (c_1 + c_2) x^3$

= $ (a_1 + a_2) ( 1 + x^2) + (c_1 + c_2) x^3$

Which does not satisfy the first axiom.

But I was checking the correction, it turns out that it actually does satisfy the first axiom and S is a subspace of P3.

I still don't get why since there was no explanation in the paper, I'm hoping someone here could enlighten me and explain what I did wrong.

Thank you.

Answer 1

You already are on the right track.

$$\underbrace{(a_1+a_2)}_a-\underbrace{(a_1+a_2)}_b=0$$

Looking at the axiom carefully you will notice that $S$ is basically

$$ \{S = a (1+x^2) + cx^3|a,c\in\mathbb{R}\} $$

Answer 2

Consider $(a_1+a_2)$ as your $a$ and $(b_1+b_2)$ as your $b$.

Then,

$a-b=(a_1+a_2)-(b_1+b_2)=(a_1-b_1)+(a_2-b_2)=0$

Which satisfies the first axiom.