Let $V$ be a finite dimensional vector space over a field $k$ and let $V^*=\Hom(V, k)$ be the dual vector space of $V$.
Let $\{v_i\}_{i=1}^n$ be a basis of $V$ and let $\{v^i\}_{i=1}^n$ be the dual basis of $V^*$. Then prove that
\[x=\sum_{i=1}^nv^i(x)v_i\] for any vector $x\in V$.

Proof.

Recall that the dual basis $\{v^i\}_{i=1}^n$ consists of vectors $v^i \in V^*$ satisfying
\[v^j(v_i)=\delta_{i,j}, \tag{*}\] where $\delta_{i,j}$ is the Kronecker delta function that is $1$ if $i=j$ and $0$ if $i\neq j$.

Let $x$ be an arbitrary vector in $V$.
Since $\{v_i\}_{i=1}^n$ is a basis of $V$, we express $x\in V$ as a linear combination of the basis. We have
\[x=\sum_{i=1}^nc_iv_i,\] where $c_i$ is a scalar (an element in $k$) for $i=1, \dots, n$.
For a fixed $j$, we have
\begin{align*}
v^j(x)&=v^j \left(\sum_{i=1}^nc_iv_i \right)\\
&=\sum_{i=1}^nc_iv^j(v_i) && \text{ by the linearity of $v_j$}\\
&=\sum _{i=1}^nc_i \delta_{i,j} && \text{ by (*)}\\
&=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
\[x=\sum_{i=1}^nv^i(x)v_i\] as required.

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