Q :Let V be a finite-dimensional vector space over a field k and let $V^\ast=Hom(V,k)$ be the dual space of V. Let $\left\lbrace v^i \right\rbrace^n_{i=1} $ be the dual basis of $V^\ast$. Then prove that for any linear functional $\sigma\in V^\ast$ $$\sigma=\sigma(v_1)v^1+\sigma(v_2)v^2+\cdots+\sigma(v_n)v^n$$
My approach :Recall that the dual basis $\left\lbrace v^i \right\rbrace^n_{i=1}$ consists of vector $v^j(v_i)=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta function that is $1 \text{ if } i= j$ and $0 \text{ if } i\neq j.$ Let x be an arbitrary vector in V. Since $\left\lbrace v^i \right\rbrace^n_{i=1} $ is a basis of V, we express $x\in V$ as a linear combination of the basis. We have $x=\sum^n_{i=1}c_iv_i$ where $c_i\in k$. For a fixed j, we have
\begin{align*}
v^j(x) &= v^j\left( \sum^n_{i=1}c_iv_i \right) \\ &=\sum^n_{i=1}c_iv^j(v_i) \qquad \text{by the linearity of }v_j \\ &=\sum^n_{i=1}c_i\delta_{ij} \qquad \text{by the definition of dual basis}\\ &=c_j
\end{align*}
Thus we have obtained $c_j=v^j(x)$ for any j. Substituting this into the linear combination of x, we have $x=\sum^n_{i=1}v^i(x)v_i-(*)$.
Then any $\sigma \in V^\ast$ let us take an action $\sigma$ in both side of $equ^n-(*)$
\begin{align*}
\sigma(x)=\sigma\left( \sum^n_{i=1}v^i(x)v_i \right)
\end{align*}
But how to get rid of the sum now. I have no idea how to approach further. Besides I am confused..am I going to right direction$?$ Any hint or solution will be appreciated .
Thanks in advance .
There are $\alpha_1,..., \alpha_n \in k$ such that $ \sigma=\alpha_1v^1+...+\alpha_nv^n.$
Then we get $\sigma(v_j)=\alpha_j$.