Linear algebra: $Ax=b$, proof

Question

$A$ is a matrix, $b$ is a vector. Show that if the equation $Ax = b$ has two distinct solutions, then it has infinitely many solutions.

Can someone help me with this through clear explanations?

Answer 1

If $x$ and $y$ are both distinct solutions then,

$Ax = b$ and $Ay = b$,

then

$Ax + Ay = A(x + y) = 2b$,

so $\frac12 (x + y)$ is a solution as well.

Answer 2

Suppose $x$ and $y$ are two distinct solutions.
Then so are $(1-\lambda)x+\lambda y$ for every $\lambda \in \mathbb{R}$ because $$A((1-\lambda)x+\lambda y) = (1-\lambda)Ax+\lambda Ay = (1-\lambda)b+\lambda b = b.$$ Hence the entire line through $x$ and $y$ solves also the given linear system. This line contains infinitely many points because $x\neq y$.

Answer 3

If $Ax_1=Ax_2=b$ and $x_1 \ne x_2$ then put

$z_t=t(x_1-x_2)+x_1$ for $t \in \mathbb K$.

Then $Az_t=b$ for all $t \in \mathbb K$.

Answer 4

Suppose $A\mathbf{x}=\mathbf{b}$ has at least two solutions, say $\mathbf{x}_1$ and $\mathbf{x}_2$, with $\mathbf{x}_1 \neq \mathbf{x}_2$.

Let $t \in \mathbb{R}$. Denote by $\mathbf{x}_0$ the difference $\mathbf{x}_2 - \mathbf{x}_1$; so $\mathbf{x}_0 \neq \mathbf{0}$. Then \begin{alignat*}{2} A(\mathbf{x}_1 + t \mathbf{x}_0) &= A\mathbf{x}_1 + A(t \mathbf{x}_0) &\qquad & \\ &= \mathbf{b} + t(A \mathbf{x}_0) &\qquad &\text{($\mathbf{x}_1$ is a solution)} \\ &= \mathbf{b} + t\bigl(A (\mathbf{x}_2 - \mathbf{x}_1) \bigr) &\qquad & \\ &= \mathbf{b} + t (A \mathbf{x}_2 - A \mathbf{x}_1) &\qquad & \\ &= \mathbf{b} + t(\mathbf{b} - \mathbf{b}) &\qquad &\text{($\mathbf{x}_1$, $\mathbf{x}_2$ are solutions)} \\ &= \mathbf{b}. &\qquad & \end{alignat*} Thus, $\mathbf{x}_1 + t \mathbf{x}_0$ is a solution. Since $t$ was arbitrary (it can take on infinitely many values) and $\mathbf{x}_0 \neq \mathbf{0}$, the solution set of $A \mathbf{x} = \mathbf{b}$ includes the infinite set $\{ \mathbf{x}_1 + t \mathbf{x}_0 \mid t \in \mathbb{R} \}$.

Therefore, there are infinitely many solutions.