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I don't understand the task from (b). Is it equivalent to : for every linear functional $f(x_{1}, ..., x_{n})=c_{1}x_{1}+...+c_{n}x_{n}$ on $F^{n}$ which satisfy $c_{1}+...+c_{n}=0$ there exists exactly one and unique functional that belongs to the dual space of $W$?

If it is so, could someone please give me a hint?

shooting-squirrel
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2 Answers2

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Yes. That is, for each $(c_1,\ldots,c_n)\in W$, define $$f_{(c_1,\ldots,c_n)}(x_1,\ldots,x_n) = c_1x_1 + \cdots + c_nx_n$$ and let $W'= \{f_{(c_1,\ldots,c_n)} : (c_1,\ldots,c_n)\in W\}$. Then $W'$ is a subspace of $(F^n)^\ast$.

Part (b) asks to prove that there exist an isomorphism between $W^\ast$ and $W'$.

Clearly $W$ and $W'$ are isomorphic. Can you take it from here?

leo
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I worked out the details from leo's hint.

Consider the sequence of functions $$W\rightarrow (F^n)^{*} \rightarrow W^{*}$$ where the first function is $$(c_1,\dots,c_n)\mapsto f_{c_1,\dots,c_n}$$ where $f_{c_1,\dots,c_n}(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ and the second function is restriction from $F^n$ to $W$

We know both $W$ and $W^{*}$ have the same dimension. Thus if we show the composition of these two functions is one-to-one then it must be an isomorphism. Suppose $(c_1,\dots,c_n)\in W\mapsto f_{c_1,\dots,c_n}=0\in W^{*}$.

Then $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)\in W$.

In other words $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)$ such that $\sum x_i=0$.

Let $\{\alpha_1,\dots,\alpha_{n-1}\}$ be the basis for $W$ from part (a) ($\alpha_i$ has one in the first component and $-1$ in the $i$-th component). Then $f_{c_1,\dots,c_n}(\alpha_i)=0$ $\forall$ $ i=1,\dots,n-1$; which implies $c_1=c_i$ $\forall$ $i=2,\dots,n$. Thus $\sum c_i=nc_1$. But $\sum c_i=0$, thus $c_1=0$. Thus $f_{c_1,\dots,c_n}$ is the zero function.

Thus the mapping $W\rightarrow W^{*}$ is a natural isomorphism. We therefore naturally identify each element in $W^{*}$ with a linear functional $f(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ where $\sum c_i=0$.

Gregory Grant
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  • Isn't it $\sum c_i=nc_1$? – Bijesh K.S Jul 16 '17 at 15:03
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    @BijeshK.S Yes I think you're right, I looked at it carefully because I remeember there was some reason I put $n-1$ but I can't figure out why now, so I guess $n$ is right. – Gregory Grant Jul 17 '17 at 00:19
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    I know this is old and I'm nitpicking, but $nc_1 = 0$ need not imply $c_1 = 0$ if $F$ has positive characteristic. But I think you can just argue that your maps sends any basis element $\alpha_i = e_1 - e_i$ to $e^1 - e^i$ (where $(e^i)$ denotes the dual basis of the canonical basis $(e_i)$), and that $(e^1 - e^i)$ forms a basis for $W^*$. – Alex Provost Jul 17 '17 at 09:07
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    @AlexProvost It's not nitpicking at all you're absolutely right. I'll fix it thanks for your suggestion. – Gregory Grant Jul 17 '17 at 13:54
  • You do not need to: $\sum c_i x_i = c\sum x_i = (nc/n)\sum x_i = ((\sum c_i)/n)\sum x_i =$ $(0/n)\sum x_i = 0$. – user0 Jan 29 '21 at 19:15