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We are asked to prove this by either contradiction, or by considering the row-echelon forms of $(\textbf{A } | \textbf{ 0})$ and $(\textbf{A } | \textbf{ b})$.

I'm not sure how to approach this.

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    $\begingroup$ If $Ax_1=b=Ax_2$ then $A(x_1-x_2)=?$ $\endgroup$
    – lulu
    Sep 3, 2018 at 20:16
  • $\begingroup$ Have you tried proving it by contradiction? What happened? $\endgroup$
    – saulspatz
    Sep 3, 2018 at 20:16
  • $\begingroup$ To elaborate on what lulu wrote, just so you understand what they suggested: suppose that it is NOT unique, so there exists $x_1$ and $x_2$ that satisfy $Ax_1 = b$ and $Ax_2 = b$. Then carry on with lulu's hint. $\endgroup$
    – anak
    Sep 3, 2018 at 20:17
  • $\begingroup$ I think there is also the case that $Ax = b$ has no solution $\endgroup$
    – Mark
    Sep 3, 2018 at 20:19
  • $\begingroup$ Note: there is no need for $Ax=b$ to have any solution at all, it just can't have more than one. Suppose that the underlying vector space is the space of infinite sequences $(x_1,x_2,\cdots)$ of real numbers and that $A(x_1,x_2,\cdots)=(0,x_1,x_2,\cdots)$. Then $Ax=0$ has only the trivial solution but there is no solution to $Ax=(1,0,\cdots)$. To be sure, the situation is different if the underlying vector space is finite dimensional. $\endgroup$
    – lulu
    Sep 3, 2018 at 20:36

4 Answers 4

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Suppose $x'$ and $x''$ both satisfy $Ax' = b$ and $Ax''=b$. Then $A(x''-x') = 0$. However, if $Ax=0$ has only the trivial solution, this implies that $(x''-x')$ must itself be 0, which implies $x''=x'$.

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Let $A x_1 = A x_2 = b$. Then $A(x_1-x_2)=0$. Thus $x_1=x_2$.

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Let me give you an alternative approach using the concept of solution sets:

Note, that for $S(A,b)=\{x\in\mathbb{R}^n\mid Ax=b\}\neq\emptyset$, we have $$S(A,b)=x+S(A,\mathbf0)\text{ for some }x\in S(A,b)$$ i.e. $S(A,b)$ is a linear shift of the null space $S(A,\mathbf0)$ by some(any!) known solution $x$ to $Ax=b$.

Thus, if $Ax=\mathbf0$ has the only solution $x=\mathbf0$, then $S(A,\mathbf0)=\{\mathbf0\}$ and followingly, we have

$$S(A,b)=x+S(A,\mathbf0)=x+\{\mathbf0\}=\{x\}$$

for some $x\in S(A,b)$(which is followingly the only one).

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    $\begingroup$ But why is there a solutions at all? $\endgroup$ Sep 3, 2018 at 20:23
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Let there be two distinct solutions,$$Ax_0=b,\\Ax_1=b.$$

Then by subtraction,

$$Ax_0-Ax_1=A(x_0-x_1)=0$$

which implies

$$x_0-x_1=0,$$ a contradiction.

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