1. Vector subspaces

I am having difficulty with vector subspaces. It would be helpful if someone could give me an answer to at least one of these two questions.

1.

Prove that $\displaystyle W_1= \{(a_1,a_2· \cdot \cdot \cdot ,a_n) \in F^n : a_1+a_2\ + \cdot \cdot \cdot a_n =0 \}$ is a subspace of F^n, but $\displaystyle W_2= \{(a_1,a_2\,\cdot \cdot \cdot a_n) \in F^n : a_1+a_2 + \cdot \cdot \cdot a_n =1 \}$ is not

2.

Is the set W = { f(x) $\displaystyle \in$ P(F): f(x)=0 or f(x) has a degree n} a subspace of P(F) if n ≥ 1? Justify your answer

2. $\displaystyle W_1= \{(a_1,a_2,...,a_n) \in F^n : a_1+a_2+...+ a_n =0 \}$

If $\displaystyle a=(a_1,a_2,...,a_n)\in W_1$ , $\displaystyle b=(b_1,b_2,...,b_n) \in W_1$ and $\displaystyle t,u \in \mathbb{F}$ then

$\displaystyle (ta+ub)_1 + (ta+ub)_2 + ... + (ta+ub)_n = (ta_1+ub_1)+(ta_2+ub_2)+...+ (ta_n+ub_n)$

$\displaystyle = t(a_1+a_2+...+a_n) + u(b_1+b_2+...+b_n)$

$\displaystyle = t\cdot0+u\cdot 0$

$\displaystyle = 0$

Therefore, $\displaystyle ta+ub \in W_1 \implies W1$ is a subspace.

3. $\displaystyle W_2= \{(a_1,a_2,...,a_n) \in F^n : a_1+a_2 +...+a_n =1 \}$

$\displaystyle W_2$ is not a vector space because $\displaystyle (0,0,...,0) \not\in W_2$. Note that $\displaystyle (0,0,...,0)$ is simply the zero vector.

4. Originally Posted by Matharch
2.[/COLOR]

Is the set W = { f(x) $\displaystyle \in$ P(F): f(x)=0 or f(x) has a degree n} a subspace of P(F) if n ≥ 1? Justify your answer[/COLOR]
[/tex]f(x)= x^2+ 1[/tex] is a polynomial of degree 2. $\displaystyle g(x)= -x^2+ 3$ is a polynomial of degree 2 also. Is there sum a polynomial of degree 2? Is the set of "polynomials of degree 2" closed under addition?

5. Thanks guys! It's so simple, I'm almost ashamed that I posted here.

6. Originally Posted by HallsofIvy
[/tex]f(x)= x^2+ 1[/tex] is a polynomial of degree 2. $\displaystyle g(x)= -x^2+ 3$ is a polynomial of degree 2 also. Is there sum a polynomial of degree 2? Is the set of "polynomials of degree 2" closed under addition?
i didn't think that the second question could be answered this way.
the "Or" made me confused,since it means union.