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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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2 Answers 2

up vote 5 down vote accepted

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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I'm slightly worried about the characteristic of the field. Will update when I figure out the confusion. –  Dylan Moreland Apr 22 '12 at 18:47
    
Not sure if the OP knows about the first isomorphism theorem... –  user38268 Apr 22 '12 at 22:54
    
@BenjaminLim: I know those Theorems. –  spohreis Apr 22 '12 at 23:26
    
Well, there are other ways to verify that this. This just seemed to be the quickest one. And I flipped through Hoffman-Kunze and it appears to me that fields are assumed to be subfields of $\mathbf C$ unless otherwise noted, so we're safe. –  Dylan Moreland Apr 22 '12 at 23:34
    
@DylanMoreland $\dim W^{\circ} \leq \dim V$. Therefore we have a surjective map $f : V \longrightarrow W^{\circ}$ with kernel $W$. Therefore by the FIT we have that $V/W \cong W^{\circ}$. How do we conclude from here that the isomorphism $(V/W)^{\ast} \cong W^{\circ}$ is natural? –  user38268 Apr 23 '12 at 0:00

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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Why are you defining a map from $W$ into $W^{\star}?$ –  spohreis Apr 20 '12 at 18:39
    
I am trying to identify $W$ as a guy sitting in $W^\ast$. So, an injectivehomomorphism does that, it renames every element in $W$ by an equivalent name in $W^\ast$. So, $W$ sits inside $W^\ast$... –  user21436 Apr 20 '12 at 18:46
    
But, why do you want to define an injective homomorphism from $W$ into $W^{\star}?$ If I understood the question correctly, we want to define an isomorphism from $W^{\star}$ onto $\{f:V\rightarrow \mathbb{F}|f(x_{1},\ldots,x_{n})=\sum_{i=1}^{n}c_{i}x_{i} \operatorname{and} \sum_{i=1}^{n}c_{i}=0\}.$ I will be back soon. –  spohreis Apr 20 '12 at 18:53
    
You definitely won't have an isomorphism. (They asked you to identify $W$ as a subspace of $W^\ast$...) You have linear functionals of the form $\underline x \mapsto \sum_{i=1}^n c_i x_i$ where $\sum c_i \neq 0$.. –  user21436 Apr 20 '12 at 18:57
    
@KannappanSampath Actually I am trying to understand why $W^{\ast}$ is not isomorphic to $W$ under your map. –  user38268 Apr 22 '12 at 14:33

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