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I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $1$ from page $111$.

If $W$ is a subset of a vector space $V$, we define $W^{0}=\{f\in V^{\star}|f(w)=0 \operatorname{for all}w\in W\}.$

Let $n$ be a positive integer and $\mathbb{F}$ be a field. Let $W$ be a set of all vectors $(x_{1},\ldots,x_{n})$ in $\mathbb{F}^{n}$ such that $x_{1}+\cdots+x_{n}=0$.

a) Prove that $W^{0}$ consists of all linear functionals $f$ of the form $f(x_{1},\ldots,x_{n})=c(\sum_{j=1}^{n}x_{j}).$

b) Show that the dual space $W^{\star}$ of $W$ can be "naturally" identified with the linear functionals $f(x_{1},\ldots,x_{n})=c_{1}x_{1}+\cdots+c_{n}x_{n}$ on $\mathbb{F}^{n}$ which satisfy $c_{1}+\cdots+c_{n}=0$.

My approach for part a) First, we note that $e_{i}-e_{j}\in W$ for all $i,j\in\{1,2,\ldots,n\}$. If $f\in W^{0},$ then we have $f(e_{i})-f(e_{j})=f(e_{i}-e_{j})=0.$ Therefore $f(e_{i})=f(e_{j})$ and we have $f(e_{i})=c$, for all $i\in \{1,2,\ldots,n\}$, where $c$ is a constant. Therefore, if $w=(x_{1},\ldots,x_{n})\in W$, then $f(x_{1},\ldots,x_{n})=c(\sum_{j=1}^{n}x_{j}).$ The converse is easy.

I was not able to solve the part b).

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Let \[ H = \{f \in (\mathbb F^n)^* \mid f(x_1, \ldots, x_n) = \sum c_ix_i \text{ and } \sum c_i = 0 \}. \] There's a map $(\mathbb F^n)^* \to W^*$ given by restriction—if you like, this is the dual $i^*$ of the inclusion $i\colon W \to \mathbb F^n$. Restrict this to a map $H \to W^*$ and show that you obtain an isomorphism. For this, you could use the fact that the kernel of $i^*$ is precisely $W^0$. What is $W^0 \cap H$?

I take the word "natural" here to mean, "Don't just identify $W$ and $H$ because they have the same dimension." The word often means "basis-free" (or functorial), but here it seems to me that we've chosen a basis by merely writing $\mathbb F^n$. In general, if I have a vector space $V$ and a subspace $W \subset V$ then the dual of $W$ is most naturally identified with the quotient $V^*/W^0$. (In the same spirit, can you find something naturally isomorphic to $(V/W)^*$?)

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    Are you referring to my parenthetical problem? I just meant that from an element of $(V/W)^*$ I get an element of $V^*$ by (pre)composing with the quotient map $V \to V/W$. This association is injective, and you just have to check that the image is right.2012-04-23
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This does not address OP's concern, but this does something irrelevant and trivial.


The part $(b)$ is largely about writing the details down:

$W=\{\underline w =(w_1,\dots,w_n) \mid \sum_{i=1}^nw_i=0\} \tag{1}$ $W^{\ast}=\{f \mid f\;\; \mbox{linear function from } W \;\;\mbox{to}\;\;\Bbb F\} \tag{2 }$

Let's also define $f_{\underline x}:W \to \Bbb F$ given by $f_{\underline x}(\underline w)=\underline x \cdot \underline w$. Let's use this to define the following map:

$\begin{align}W &\to W^\ast \\ \underline x&\mapsto f_{\underline x}\end{align}$

Proving that the above map is an injective homomorphism identifies $W$ with $W^\ast$. Since, $\dim W=\dim W^\ast$, we also have that this map is an isomorphism.

But OP wants an altogether different isomorphism, so, I GOOFED IT UP.

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    @KannappanSampath I think the problem is not about identifying $W$ with $W^{\ast}$,but rather $W^{\ast}$ with the object the OP defined above.2012-04-22