Linear equation systems have 0 ∨ 1 ∨ ∞ solutions

Show that a system of linear equations either has no solution, or exactly one solution, or infinitely many solutions.

Let $A\in\mathbb{R}^{m\times n}$, $b\in\mathbb{R}^m$, and consider the linear system

$$ Ax=b. $$

Let $S=\{x\in\mathbb{R}^n:Ax=b\}$ be its solution set. There are three possibilities:

---

1. No solution. The system is inconsistent (e.g., row-reduction of the augmented matrix $[A\mid b]$ produces a row $[0\ \cdots\ 0\mid c]$ with $c\neq 0$). Then $S=\varnothing$.

---

2. At least one solution. Suppose $x_0$ is a solution: $Ax_0=b$. Define the homogeneous solution space

$$ H=\{h\in\mathbb{R}^n:Ah=0\} $$

(which is a subspace of $\mathbb{R}^n$). Then for any $h\in H$, $A(x_0+h)=Ax_0+Ah=b+0=b$, so $x_0+H\subseteq S$. Conversely, if $x\in S$, then $A(x-x_0)=Ax-Ax_0=b-b=0$, so $x-x_0\in H$ and $x=x_0+h$ for some $h\in H$. Hence

$$ S = x_0 + H, $$

i.e., the solution set is an affine subspace parallel to the homogeneous solution space.

Now:

• If $H=\{0\}$, then $S=\{x_0\}$: exactly one solution.
• If $H$ contains a nonzero vector $h\ne 0$, then for each real scalar $t$, $x_0+th\in S$. If $t_1\ne t_2$, then $(x_0+t_1h)-(x_0+t_2h)=(t_1-t_2)h\ne 0$, so these are all distinct. Thus there are infinitely many solutions.

---

Conclusion: Over $\mathbb{R}$ (or any infinite field), a linear system has either no solution, exactly one solution, or infinitely many solutions—no other cardinalities occur.

Example:

• $x+y=1,\ x+y=2$ is inconsistent ⇒ no solution.
• $x+y=1,\ x-y=0$ has a unique solution $x=y=\tfrac12$.
• $x+y=2,\ 2x+2y=4$ is consistent but dependent: $S=\{(t,2-t):t\in\mathbb{R}\}$ ⇒ infinitely many solutions.

(Note: over a finite field, the “infinitely many” case becomes “more than one” but still finite; the trichotomy above assumes the underlying field is infinite, as in $\mathbb{R}$.)