Infinite dimensional vector spaces

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 $M$.

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 $$m:=v-\sum_{l=2}^r f_l(v)m_l.$$
Each of the $f_i$ has a 1-codimensional kernel, and the intersection of all those kernels, for $i=2,\cdots,r$ is a $(r-1)$-codimensional subspace, which is not empty, if $M$ is infinite dimensional; let us call it $W$. We would like $m$ to be in $W$, so that $f_i(m)=0$ for $i\geqslant 2$. Can we find one such that $f_1(m)\neq 0$? Yes, otherwise, $W\subset ker(f_1)$; making the quotient we have $\bar{f}_1,\cdots,\bar{f}_r\in(M/W)^*$, which cannot be linear independent (because $(M/W)^*$ has dimension $r-1$), but any linear dependency between the $\bar{f}_i$ leads to a linear dependency between the $f_i$ when coming back to $M$.
• why $W$ is $r-1$ codimensional subspace? – mathuser77 Dec 10 '14 at 9:22
Take $r$ linear independent vectors $v_1,\dots,v_r \in M$, define $A=(A_{i,j})_{i,j=1}^r=(f_i(v_j))_{i,j=1}^r$, $\lambda=A^{-1}e_1$ and $m=\sum_{i=1}^r \lambda_iv_i.$ Then $$f_i(m)=\sum_{j=1}^r \lambda_j f_i(v_j)=\sum_{i=1}^r A_{ij}\lambda_j=e_1(j)=\delta_{1j}.$$
• what is the element e_1? and why $A$ is invertible? – mathuser77 Dec 10 '14 at 9:20
• $e_1=(1,0,\dots,0)$. Unfortunately $A$ does not have to be invertible (it could be zero for example) so the above is not a solution. – user35593 Dec 10 '14 at 9:51