Let $V$ be a real vector space of dimension $n$. It is well-known that if $v_1,\dots,v_k$ are linearly independent vectors in $V$ (of course $k<n$), then there are $f_1,\dots,f_k\in V^*$, where $V^*$ is the dual space of $V$, such that $f_i(v_j)=\delta_{ij}$, and in this case $f_1,\dots,f_n$ are linearly independent.
Is the converse of this also true? That is, suppose $f_1,\dots,f_k\in V^*$ are linearly independent. Then are there $v_1,\dots,v_k\in V$, which are linearly independent, such that $f_i(v_j)=\delta_{ij}$? It is clear that if $f_i(v_j)=\delta_{ij}$, then the $v_i$'s are linaerly independent, so we only need to show that there exists $v_1,\dots,v_k$ such that $f_i(v_j)=\delta_{ij}$.
Best Answer
Yes, the converse is true.
Consider the map$$\begin{array}{rccc}\Psi\colon&V&\longrightarrow&(V^*)^*\\&v&\mapsto&\left(\begin{array}{ccc}V^*&\longrightarrow&k\\\alpha&\mapsto&\alpha(v)\end{array}\right),\end{array}$$where $k$ is the field that you're working with. Then $\Psi$ is injective and, since $V$ is finite-dimensional, it is then an isomorphism. So, take $\alpha_1,\ldots,\alpha_n\in(V^*)^*$ such that $\alpha_i(f_j)=\delta_{ij}$ and let $v_i=\Psi^{-1}(\alpha_i)$.