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?
Best Answer
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 $\mathbf{c}=\begin{pmatrix}c_1 & c_2 & \cdots & c_n\end{pmatrix}^\mathrm{T}$.