Suppose $Ax=b$, where $A$ is of dimension $q\times p$ with $q<p$, $rank(A)=q$, $b$ is $q \times 1$. Let $\mathcal{X}=\{x:Ax=b\}$ be the solution set of this system. How to rigorously prove that any solution $x^{*}\in \mathcal{X}$ could be written as $x^{*}=B\pi_{free}+w$ for some free parameter vector $\pi_{free}$ of dimension $(p-q)\times 1$, $B$ of dimension $p\times (p-q)$, $w$ of dimension $p\times 1$? Example: $x=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}$, $A=\begin{bmatrix}1&0&-1\\0&1&-1\end{bmatrix}$, $b=\begin{bmatrix}1\\2\end{bmatrix}$, then any solution to this system could be written as $x^{*}=\begin{bmatrix}1\\1\\1\end{bmatrix}x_{free}+\begin{bmatrix}1\\2\\0\end{bmatrix}$ with $B=\begin{bmatrix}1\\1\\1\end{bmatrix}$, $x_{free}=x_{3}$, and $w=\begin{bmatrix}1\\2\\0\end{bmatrix}$. Thank you very much!

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    $\begingroup$ To get started, let $w$ be a particular solution of $Ax=b$. Note that if $x^*$ is also a solution to $Ax=b$, then $x^* - w$ is an element of the null space of $A$. (And what is the dimension of the null space of $A$?) $\endgroup$ – littleO Aug 17 at 20:24
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    $\begingroup$ Ah, I see it now. This is very helpful. Thank you so much! $\endgroup$ – TD888 Aug 17 at 23:49

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