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 \(AX = 0\).
Solution
The linear map \[ T:F^n\longrightarrow F^s, \qquad T(X)=AX, \] has rank at most \(s\). Since \(s<n\), rank-nullity gives \[ \dim\ker T=n-\operatorname{rank}T\ge n-s>0. \] Therefore \(\ker T\) contains a nonzero vector \(X\). For that vector, \[ AX=T(X)=0. \]