Here is the question : If $f_{1},\dots,f_{p}$ are linear functionals on an $n$-dimensional vector space, $X$, where $p < n$, show that there is a vector $x \neq 0$ in $X$ such that $f_{1}(x) = \cdots =f_{p}(x) = 0$.
I really want to prove this by contradiction. I want to consider this function $x \mapsto (f_{1}(x),..., f_{p}(x))$. I know for the dual basis, $X^*$, there are Kroneker delta's but I do not know how to apply them properly. Is this the way to go?
Thank you very much!
Let $\mathbb{k}$ be the field. Let $\{e_1,...,e_n \}$ be a basis of $X$. For every $i \in \{1,...,n\}$ let $\epsilon_i$ be the unique linear functional $X \to \mathbb{k}$ such that $\epsilon_i(e_j)=\delta_{i,j}$. Then as you probably know $\{\epsilon_1,...,\epsilon_n \}$ is a basis of $X^*$.
Now, without loss of generality we can assume that the $f_i$'s are linearly independent in $X^*$, otherwise you write someone of them as linear combination of the other ones and then you just consider these ones.
Then we can complete the linearly independent set $\{f_1,...,f_p\}$ to a basis $\{f_1,...,f_n\}$ of $X^*$. As you probably know (Q), there is unique a basis $\{e_1,...,e_n \}$ of $X$ such that $f_i=\epsilon_i$ for every $i \in \{1,...,n\}$.
Hence it is the case that $f_i(e_{p+1})=\epsilon_i(e_{p+1})=0$ for every $i \in \{1,...,p\}$ and of course $e_{p+1}\neq 0$.
How to prove Q. Let us consider the usual isomorphism $\alpha \colon X \to X^{**}$ such that, for every $x \in X$, it is the case that $\alpha x$ is the linear functional $X^* \to \mathbb{K}$ such that $(\alpha x)(f)=f(x)$ for every $f \in X^*$. As $\{f_1,...,f_n \}$ is a basis of $X^*$, it is the case that $\{\epsilon'_1,...,\epsilon'_n \}$ is a basis of $X^{**}$, being $\epsilon'_i$ the unique linear functional $X^* \to \mathbb{k}$ such that $\epsilon'_i(f_j)=\delta_{i,j}$. Moreover, being $\alpha$ an isomorphism $X \to X^{**}$, it is the case that $\{e_1:=\alpha^{-1}\epsilon'_1,...,e_n:=\alpha^{-1}\epsilon'_n \}$ is a basis of $X$.
Verify that the basis $\{\epsilon_1,...,\epsilon_n \}$ of $X^*$ given by the basis $\{e_1:=\alpha^{-1}\epsilon'_1,...,e_n:=\alpha^{-1}\epsilon'_n \}$ of $X$ is precisely the one that we were looking for.
The picture here is that each nonzero linear functional on an $n$-dimensional vector space has an a kernel that is at least $(n-1)$-dimensional. Since you have $p<n$ functionals, the intersection of their kernels is nontrivial. (In addition to Federico's answer, another way of seeing this is by choosing coordinates, observing that to be in the kernel of one of the $f_k$ you must be perpendicular to its gradient vector $\nabla f_k$, and then noting those gradients span a $p$-dimensional subspace whose complement must be nontrivial.)
Another way of putting it: "$x$ such that $f_i(x) = 0$ for all $i = 1,\ldots,p$" is actually a system of $p$ linear equations. Since you have $n$ variables and $p$ equations, the system is underdetermined and has nontrivial solutions. You can formalize this by choosing a basis and writing down a matrix equation.
A third way of seeing this: As you suggest, consider the map $F: X \to \mathbb{R}^p$ defined by $F(x) = (f_1(x), \ldots, f_p(x))$. By the rank-nullity theorem, $F$ has a nontrivial kernel. (Of course, first check that this lemma is not in your proof of the rank-nullity theorem so you can be sure this is not a circular argument.)
Your map $\phi: x\mapsto (f_1(x),\dots ,f_p(x))$ is linear, so
$\mathbb{R}^n\cong Im(\phi)\oplus ker(\phi)$
But $p< n$ so
$dim(Im(\phi))\leq dim(\mathbb{R}^p)=p<n$
Thus $dim(ker(\phi))\geq 1$ that means $ker(\phi)=\bigcap_{k}ker(f_k)\neq \{0\}$
So there exists $x\neq 0$ such that
$x\in \bigcap_{k}ker(f_k)$ that means
$f_1(x)=f_2(x)=\dots=f_p(x)=0$
You can do a similar question also in the infinite dimensional case.
Let $\Lambda$ be a set of functionals of an infinite normed vector space $V$ such that
$ cl(\Lambda) \neq V^*$. Is it true that there exists $x\in V$, $x\neq 0$, such that $f(x)=0$ for each $f\in \Lambda$?
By a corollary of Hahn-Banach theorem, applied on the dual space of $V$, there exists a non-zero functional $F$ of $V^{**}$ such that is $0$ on $\Lambda$.
Now if the space $V$ is reflexive, so the canonical injection $j: V\to V^{**}$ is surjective, then there exists a point $x\neq 0$ such that $F=j(x)$.
Thus for each $f\in \Lambda$ we have
$f(x)=j(x)(f)=F(f)=0$
