TITLE: If $f_1,...,f_p$ are linear functionals on an $n$-dimensional vector space, show that there is $x \neq 0$ such that $f_{1}(x) = \cdots =f_{p}(x) = 0$
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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.)