Proving equality in dual space

Question

Let $ V $ a finite dimensional vector space and $ V^* $ the dual space. Further let be $ B=(v_1,...,v_n) $ a base of $ V $ and $ B^*=(v_1^*,...,v_n^*) $ a dual base of $ V^*$. Show:

For all $ v,w\in V $ is $ v=w $ if and only if $ f(v)=f(w) $ for all $ f\in V^* $.

My idea:

At first we have $ f=f(v_1)\cdot v_1^*+...+f(v_n)\cdot v_n^* $. Then we get with $ v=\sum\limits_{i=1}^n a_i\cdot v_i $ and $ w=\sum\limits_{i=1}^n b_i\cdot v_i $ for all $ v,w\in V $

$ f(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n a_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot a_j\\=\sum\limits_{j=1}^n f(v_j)\cdot b_j\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n b_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(w)\\=f(w) $

Now I want to show that $ a_j=b_j $ for all $ j=1,...,n $ but I don't know how.

Answer 1

This seems too complicated. Just note that $v_i^*(\sum_j a_j v_j) = v_i^*(\sum_j b_j v_j)$ implies $a_i=b_i$.

Answer 2

It can be useful to note that if $\mathcal{B}= (v_{1}, \ldots, v_{n})$ is a basis for $V$ and $\mathcal{B}^{*}= (\varphi_{1}, \ldots, \varphi_{n})$ is the corresponding dual basis, then for any $v \in V$,

$$ v = \varphi_{1}(v) v_{1} + \ldots + \varphi_{n}(v)v_{n}, $$

so $\mathcal{B}^{*}$ gives us the coordinates of any vector $v \in V$ with respect to the basis $\mathcal{B}$.

Answer 3

Choose each $v^*_i$ and apply $v^*_i(v)=v^*_i(w)$ then $a_i=b_i$, so $v=w$.