# Proof of subspace in $V_n$ and determination of basis over $F$

I am having trouble with this homework problem:

Let $W = \{\alpha_0 + \alpha_1x + \cdots + \alpha_{n-1}x^{n-1} \in F[x] \mid \alpha_0+\alpha_1+\cdots+\alpha_{n-1} = 0 \}$. Show that $W$ is a subspace of $V_n$ and find a basis of $W$ over $F$.

I proved that $W$ forms a subspace by taking $w_1,w_2 \in W$ and $a_1,a_2 \in F$ and showing that $a_1w_1+a_2w_2 \in W$.

I'm having trouble finding a basis of $W$. By definition, a basis of $W$ consists of a subset $S \subset W$ whose vectors are linearly independent and span the entire space $W$.

I feel like I'm thinking about this question wrong. My initial feeling is to take vectors $\{w_1,w_2,\cdots,w_{n-1}\}$ in $W$ and try to show that they are independent and span $W$, but I'm not sure if this is correct.

Please let me know if I'm on the right track. I prefer hints over full solutions.

• By the way, $W$ is the kernel of the linear map $f\mapsto f(1)$ from the space of degree $\le n-1$ polynomials to $F$. So if you already know these facts (i.e. that the latter is a vector space and the former is a linear map), you get the first part with "no" effort. – Hagen von Eitzen Jan 25 '14 at 21:25

You are correct with your method (it's the general method). To find such vectors, note that you can "see" $W$ as $F^n$ (the set of $n$-tuples of elements of $F$) using the conditions that define $W$.
If you had to find a basis of the larger space $$\{\,\alpha_0+\ldots +\alpha_{n-1}x^{n-1}\in F[x]\,\}$$ the choice of $n$ (instead of $n-1$) basis vectors should be obvious. None of these is in $W$, though (at least if you took the really obvious base). Drop one vector from this base (so that you have the correct numbre of $n-1$ vectors left) and use the dropped vector to "adjust" the remaining vectors to lie in $W$.