Jtdylktuy

$W={f(x)in P(mathbb R) colon f(x)=0 text{ or } f(x)text{ has degree }5}$, $V=P(mathbb R)$

I'm really stuck on proving this question. I know that the first axioms stating that $0$ must be an element of $W$ is held, however I'm not sure how to prove closure under addition or scalar.

I've tried using arbitrary polynomials $a$, $b$ and letting them be an element of $W$. I then have:

$$ax^5+bx^5=(a+b)x^5$$

And I'm not sure how to prove that this addition is an element of W. I know intuitively it makes sense that it's an element of W, but I'm just not sure how to proof it mathematically.

Any help would be greatly appreciated!

To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.

Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.

Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."