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I am not getting even an intuition as how to solve this problem. Please help me with a solution.

Let $n$ be a positive integer and $F$ a field. Let $W$ be the set of all vectors $(x_1, \dots , x_n)$ in $F^n$ such that $x_1+\dots +x_n =0$.

$1)$ Prove that $W^0$ (annihilator of $W$) consist of all linear functionals $f$ of the form $$f(x_1, \dots , x_n) = c \sum _{j=1}^n x_j.$$

$2)$ Show that the dual space $W^*$ of $W$ can be naturally identified with the linear functionals of the form $$f(x_1, \dots , x_n) =c_1x_1 +\dots +c_nx_n$$ on $F^n$ which satisfy $c_1+\dots +c_n=0$.

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    $\begingroup$ What is the def'n of $W^0$? $\endgroup$ – DanielWainfleet Jan 16 '16 at 7:50
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Let's look like this.

1) A functional $f$ is in $W^\circ$ if and only if $f(x)=0$ whenever $x_1+\cdots+x_n=0$. Define a functional $G$ by $$g(x)=x_1+\cdots+x_n.$$ Then the kernel of the functional $g$ is contained in the kernel of the functional $f$. With symbols this means $\ker g\subseteq \ker f$. We claim that there exists a scalar $c$ such that $g=cf$ and the proof will be finished.

Write $F^n=\ker f\oplus Fe$ where $f(e)=1.$ Then, if $x=y+\lambda e$ for some $y\in \ker f$ and $\lambda \in F$, we have $$g(x)= g(y)+\lambda f(e)=\lambda g(e)=g(e) \cdot f(x).$$ Now you have $c=g(e)$.

2) If you write $x\in W$ as $x=x_1e_1+\cdots+x_ne_n$, then apply $x_n=-x-\ldots-x_{n-1}$ to get $$x=x_1(e_1-e_n)+\cdots+x_{n-1}(e_{n-1}-e_n).$$ If $f\in W^*$, then $$f(x)=x_1 c_1+\cdots x_{n-1}c_{n-1}$$ where $c_{i}=f(e_i-e_n).$ As you see the dimension of $W^*$ is $n-1$, so that the basis of the dual of $F^n$ consists of basis for $W^*$ and an additional functional $g$ defined by $g(x_1,\ldots,x_n)=-(c_1+\cdots+c_{n-1})x_n.$ Since $(F^n)^*=W^*\oplus Fg$ by the decomposition above every functional $f\in W^*$ is of desired form. I hope this helps.

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Hint. First prove that $$ \dim W^0=\dim F^n-\dim W $$ Then prove that $\dim W=n-1$ so $$ \dim W^0=1 $$ Do you see how this proves (1)?

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  • $\begingroup$ I am getting the notion, but still am not being able to write things down logically. Can you help me further? (I would suggest to write down a complete answer and also the same for the second part.) $\endgroup$ – Qwerty Jan 16 '16 at 9:07
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    $\begingroup$ @Qwerty: I would suggest that you show some work on your part. $\endgroup$ – Daniel R. Collins Jan 20 '16 at 16:16
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Put $V = F^n$. By definition, $W^0 = \{f\in V^*:\, f\vert W = 0\}$. The inclusion $W \to W\oplus W^\perp = V$ induces an isomorphism $W^0 \oplus W^* \to V^*$, given by $(f, g) \to f + g$. Hence $$\dim W^0 = \dim V - \dim W^* = \dim V - \dim W = 1.$$ The function $f(x) = x_1 + \cdots + x_n$ clearly lies in $W^0$, so it spans it. Part (2) is just the definition of dual space.

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