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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.

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2 Answers 2

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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

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

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  • $\begingroup$ Could you explain to me why the rule a - b = 0 is important in this case? Because in another case the subset was W = {a,b,c,d | ab=0} to see if it is a subspace of $R^4$, can I use the ab=0 to disprove one of the axioms? $\endgroup$
    – Kode Ch
    Apr 14, 2018 at 16:01
  • $\begingroup$ Of course. The rule(s) used to define the set is ALWAYS important. These are the things that always stay true about that particular set since it's its definition. The axioms are for subspaces but these sets may or may not be subspaces so the sets can be proven to follow or not follow the axioms of subspaces using their defining definitions. $\endgroup$ Apr 14, 2018 at 16:07
  • $\begingroup$ @KodeCh in the example in your comment just now, you appear to be multiplying $a$ and $b$, which will make it not a subspace (the condition is not linear). In the example in your original question though it subtracting $b$ from $a$, which is a linear condition and will allow it to be a linear subspace. $\endgroup$
    – JMoravitz
    Apr 14, 2018 at 16:08
  • $\begingroup$ That makes perfect sense. Thank you very much. $\endgroup$
    – Kode Ch
    Apr 14, 2018 at 16:08
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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.

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  • $\begingroup$ So is there actually a need to assume two polynomials and work on them? Can I just use a - b=0 with axioms 1 & 6 to prove that S is a subspace? $\endgroup$
    – Kode Ch
    Apr 14, 2018 at 16:04
  • $\begingroup$ Notice that you already knows that P3 is a vector space and you need to prove that S is a linear subspace. That means you only need to show 3 axioms. Check this link $\endgroup$
    – bp7070
    Apr 14, 2018 at 16:18

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