If $f_1,\dots,f_k\in V^*$ are linearly independent, then there are $v_1,\dots,v_k\in V$ such that $f_i(v_j)=\delta_{ij}$?

Question

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 $f_i(v_j)=\delta_{ij}$.

Answer 1

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 $v_i=\Psi^{-1}(\alpha_i)$.

Answer 2

First prove the standard fact: if $g(v) = 0$ whenever $f_1(v) = \cdots = f_l(v) =0$, then $g= \sum a_i f_i$ for some $a_1$, $\ldots$, $a_l$. Proof: define the map $(f_1(v), \ldots, f_l(v)) \mapsto g(v)$, it is well defined, linear, and extends to a linear map from $\mathbb{K}^l$ to $\mathbb{K}$, so of form $(x_1, \ldots, x_l) \mapsto \sum a_s x_s$.

Now since every $f_k$ is not dependent on the other $f_l$'s, there exists $v_k$ such that $f_k(v_k) \ne 0$ and $f_s(v_k)=0$ for all $s\ne k$. Now multiply $v_k$ by constant to get $f_k(v_k) = 1$.