The number of solutions to a system of linear equations Can anyone suggest a formal proof that a system of linear equations can have no solution, one solution or infinitely many solutions? 

Consider a linear system in $\mathrm x \in \mathbb R^n$
$$\mathrm A \mathrm x = \mathrm b$$
where $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm b \in \mathbb R^m$ are given. Suppose that the system is feasible and that $\mathrm x^{(1)}$ and $\mathrm x^{(2)}$ are two solutions. Hence, $\mathrm A \mathrm x^{(1)} = \mathrm b$ and $\mathrm A \mathrm x^{(2)} = \mathrm b$. Subtracting these two, we obtain
$$\mathrm A \mathrm x^{(1)} - \mathrm A \mathrm x^{(2)} = \mathrm b - \mathrm b = 0_m$$
or,
$$\mathrm A (\mathrm x^{(1)} - \mathrm x^{(2)}) = 0_m$$
Hence, $\mathrm x^{(1)} - \mathrm x^{(2)}$ is in the null space of $\mathrm A$. If


*

*the null space is trivial (i.e., it contains only $0_n$), then $\mathrm x^{(1)} - \mathrm x^{(2)} = 0_n$, or, $\mathrm x^{(1)} = \mathrm x^{(2)}$.

*the null space is nontrivial (i.e., it is not $0$-dimensional), then any affine combination of $\mathrm x^{(1)}$ and $\mathrm x^{(2)}$ is also a solution to the linear system, for the following holds for all $\gamma \in \mathbb R$
$$\mathrm A (\gamma \mathrm x^{(1)} + (1-\gamma) \mathrm x^{(2)}) = \gamma \mathrm A \mathrm x^{(1)} + (1-\gamma) \mathrm A \mathrm x^{(2)} = \gamma \mathrm b + (1-\gamma) \mathrm b = \mathrm b$$
Thus, if a linear system is feasible, it either has one solution, or it has infinitely many.
