Chapter 2 Sec. 2.6 Hoffman Kunze Linear Algebra exercise 1 
Let $s<n$ and $A$ an $s \times n$ matrix with entries in the field $F$.
  Use theorem 4(not its proof) to show that there is a non-zero $X$ in
  $F^{n \times 1}$ such that $AX=0$.

Theorem 4:

Let $V$ be a vector space which is spanned by a finite set of vectors $a_{1},a_{2},\ldots,a_{n}$. Then any independent set of vectors in $V$
  is finite and contains no more than $n$ elements.

The problem is straightforward, I think, $AX=0$ denotes a system that is equivalent to a system denoted by $BX=0$, where $B$ is exactly as $A$ except that it has $s-n$ zero rows at the bottom. $B$'s RREF has at least one zero row at the bottom and as a consequence at least one variable will be free.
How to solve this using theorem 4?
 A: Note that the columns of the $s\times n$ matrix $A$ resides in $\mathbb{F}^s$. By theorem $4$, it follows that the set of columns of $A$ is linearly dependent (why?). Therefore there is some non-trivial linear combination of the columns, which I denote with $\mathbf{a}_i$, which sum to $\mathbf{0}$, i.e. 
$$c_1\mathbf{a}_1 + \cdots + c_n\mathbf{a}_n = \mathbf{0}$$
for $c_i\in\mathbb{F}$. It follows that we must have 
$$A\mathbf{c} = \mathbf{0}$$
where $\mathbf{c}=\begin{pmatrix}c_1 & c_2 & \cdots & c_n\end{pmatrix}^\mathrm{T}$. 
A: Matrix $A$ defines a map from $n$-dimensional vector space $V$ to $s$-dimensional $U$ with $n>s$. Hence the image of any $n$ independent vectors from $V$ must be dependent in $U$ (by theorem 4). That exactly means $AX=0$ for some $X$.
To be more precise, if your basis in $V$ is denoted $e_1,\dots,e_n$, then theorem 4 says that vectors $Ae_1,\dots, Ae_n\in U$ must be dependent since $U$ is $s$-dimensional and $s<n$. This means that $c_1Ae_1+\dots+c_nAe_n=0$ for some constants $c_i$, not all of them are zero. But this is equivalent to $AX=0$ with $X=c_1e_1+\dots c_ne_n$.
