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

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 $$k), then there are $$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 $$f_i(v_j)=\delta_{ij}$$.

• Finite dimensional spaces are reflexive: $V= V^{**}$. So yes, apply the first result to $V^*$ and it’s dual $V$. May 2 '20 at 6:21

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 $$v_i=\Psi^{-1}(\alpha_i)$$.
• I suppose you mean the map $V^*\longrightarrow k$ is $\alpha\mapsto\alpha(v)$ May 2 '20 at 6:31
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 $$f_k(v_k) = 1$$.