1
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Finite dimensional spaces are reflexive: $V= V^{**}$. So yes, apply the first result to $V^*$ and it’s dual $V$. $\endgroup$
    – User8128
    May 2 '20 at 6:21
3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ I suppose you mean the map $V^*\longrightarrow k$ is $\alpha\mapsto\alpha(v)$ $\endgroup$ May 2 '20 at 6:31
  • 1
    $\begingroup$ @MirafromEarth Sure! I've edited my answer. Thank you. $\endgroup$ May 2 '20 at 7:01
0
$\begingroup$

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 $f_k(v_k) = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.