This question might seem elementary but I cannot answer it.


Let M be an infinite dimensional vector space and $f\_1, f\_2, \cdots , f\_r \in M^\*$ be a set of linear independent vectors with $r \geq 2$. Here $M^\*:=Hom\_K(M, K)$ is the linear dual of $M$.


Do there exist $m \in M$ such that $f\_1(m)\neq 0$ but $f\_i(m)=0$ for any $i\geq 2$?


Of course the statement is obviously true for finite dimensional vector spaces $M$.


I thought about completing $\{f\_1, f\_2, \cdots, f\_r\}$ to a basis of $M^\*$ and then taking the dual basis with respect to it. But this new basis lies in $M^{\*\*}$
 which is larger then $M$.



Infinite dimensional vector spaces
You can find $m$ by Induction on $r$. For $k\in \{2,\dots,r\}$ let $m\_k$ be such that $f\_l(m\_k)=\delta\_{lk}$ for $l\in \{2,\dots,r\}$. By linear independce of $f\_1,f\_2,\dots,f\_r$ there exist $v \in M$ with $f\_1(v)-\sum\_{l=2}^r f\_1(m\_l)f\_l(v)=1$. Define now
$$m:=v-\sum\_{l=2}^r f\_l(v)m\_l.$$


