Non-zero functional on a field of characteristic zero.

Question

I noticed that this question has been addressed to varying degrees here, here, and here. The purpose of my (re)-asking it is because I wanted to get a proof that did not involve induction and a solution that was complete.

The question:

Let $F$ be a field of characteristic zero and let $V$ be a finite-dimensional vector space over $F$. If $a_1,...,a_m$ are finitely many vectors in $V$, each different from the zero vector, prove that there is a linear functional $f$ on $V$ such that $f(a_i)\neq 0$, $i=1,...,m$.

My approach is as follows. Let $W=\text{span}\{a_1,...,a_n\}.$ Let $B=\{\beta_1,...\beta_k\}$ be a basis for $W$. For all $\alpha \in W$ we have $\alpha=c_1\beta_1+...c_k\beta_k$. Then for $f(\alpha)$ we have:

$$f(\alpha)=f(c_1\beta_1+...c_k\beta_k)=c_1f_1(\beta_1)+...+c_kf_k(\beta_k).$$

The right most side is non-zero since $F$ has characteristic $0$. Hence there exists a linear functional $f$ such that $f(a_i)\neq 0$ for all $i$.

Is this approach correct?

Answer 1

I don't understand your approach (in particular, what are the $(f_k)$ ?).

Let $dim(W)=k$, $a_1,\cdots,a_k$ be a basis of $W$ and $U$ be a complementary of $W$ in $V$. We construct a linear function $f:V\rightarrow F$ s.t. for every $i$, $f(a_i)\not=0$, as follows.

$f$ is uniquely defined by, for every $j\leq k$, $f(a_j)=u_j\not= 0$ and $f_{|U}=0$. The condition to achieve is: for every $p>k$, $f(a_p)\not= 0$.

If $a_p=\sum_{j\leq k}c_{p,j}a_j\not= 0$, then necessarily $f(a_p)=\sum_{j\leq k}c_{p,j}u_j$. Then the question is

Knowing the given $(c_{p,j})$, how to find non-zero $(u_j)_{j\leq k}$ s.t. for every $p\in (k,m]$, $\sum_{j\leq k}c_{p,j}u_j\not= 0$ ?

Now we work in $F^k$. Let $u=[u_1,\cdots,u_k]^T\in F^k$ and $\delta_p=[c_{p,1},\cdots,c_{p,k}]^T\not= 0$; thus, the condition is $u\notin \cup_{p\in (k,m]} {\delta_p}^{\perp}$. Note that ${\delta_p}^{\perp}$ is a vector space of dimension $k-1$, that is a Zariski closed proper subset of $F^k$ ($F$ has characteristic $0$). Consequently, $\cup_{p\in (k,m]} {\delta_p}^{\perp}$ is a Zariski closed proper subset of $F^k$ and finally, $F^k\setminus \cup_{p\in (k,m]} {\delta_p}^{\perp}$ is Zariski open dense.

In fact, it has been shown that $\{f\in V^*;f(a_i)\not=0\}$ is dense in $V^*$ (we may define $f$ on $U$ as we want).

Answer 2

Claim: Let $V$ be a finite dimensional $K-$vector space, with $\text{char}(K) = 0.$ Let $v _1, \ldots, v _r$ be nonzero vectors. Then there is a $\varphi \in V ^{\ast}$ which is nonzero at each $v _j.$
Pf: Fix a basis $(e _1, \ldots, e _n)$ of $V.$ Expressing the $v _j$s in terms of this basis, ${ (v _1, \ldots, v _r) = (e _1, \ldots, e _n) \underbrace{[X _1, \ldots, X _r]} _{n \times r \text{ matrix } X} }$ where columns of $X$ are all nonzero.
Now for any $T \in V ^{\ast},$ ${ (T(v _1), \ldots, T(v _r)) = (T(e _1), \ldots, T(e _n)) X}.$
It suffices to find ${a _1, \ldots, a _n \in K}$ such that row $(a _1, \ldots, a _n) X$ has all entries nonzero.

Because then taking ${ (\varphi(e _1), \ldots, \varphi(e _n) ) := (a _1, \ldots, a _n)}$ would work.

That is, we want to pick an element ${ (a _1, \ldots, a _n) ^t \in K ^n \setminus (S _1 \cup \ldots \cup S _r)},$ where subspace ${ S _j := \lbrace (t _1, \ldots, t _n) ^t \in K ^n : (t _1, \ldots, t _n)X _j = 0 \rbrace }.$
Since each $X _j$ is nonzero, each $S _j$ is proper. Since $\text{char}(K) = 0,$ $K ^n$ isn't a union of finitely many proper subspaces. So such an element can be picked, as needed.

---

The claim fails when $\text{char}(K) = 0$ is dropped. Consider a $\mathbb{Z} /{2 \mathbb{Z}}$ vector space $V$ with basis $(e _1, e _2),$ and vectors ${ v _1 = e _1; v _2 = e _2; v _3 = e _1 + e _2 }.$ There cant be an $f \in V ^{\ast}$ which is nonzero on all these $v _j$s (if there were, $f(e _1)$ and $f(e _2)$ must be $1,$ but then $f(e _1 + e _2) = 1+1 = 0,$ absurd).