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I'm trying to prove that the dual of a finite dimensional vector space $V$ separates points.

Let ($v_1,v_2,...,v_n)$ be a basis. Take vectors $w_1,w_2$ in $V$ with $w_1\neq w_2$.

Suppose $w_1=\displaystyle\sum_{i=1}^n\alpha_1 v_i$ and $w_2=\displaystyle\sum_{i=1}^n\beta_1 v_i$. Since $w_1\neq w_2$, $\alpha_1\neq \beta_i$ for some $i$. Take the functional $f_i\in V^*$ given by $f_i(v_i)=1$ and $f_i(v_j)=0$ if $i\neq j$. Then $f_i(w_1)=\alpha_i\neq\beta_i=f_i(w_2)$.

Is this proof correct? I know that for infinite dimensional normed linear spaces, the proof of 'point-separation' follows from the Hahn-Banach theorem. I was trying to figure out a similar argument for finite dimensions, but ended up with this proof. Is there some mistake in it that I missed?

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  • $\begingroup$ Looks correct, but some of your $1$s should be $i$s. Yeah, the theorem is easy in the finite-dimensional case; I am not sure if it is worth imitating the infinite-dimensional proofs, since they are probably heavily burdened by topological considerations. $\endgroup$ Sep 20 '16 at 17:53
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A "coordinate-free" proof:

Let $v\neq w$. Then $(v-w)\cdot(v-w)\neq 0$, that is, $(v-w)\cdot v \neq (v-w)\cdot w$.

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