# Prove that if a linear system has infinitely many solutions, then any solution could be written as a linear function of free parameters, thanks.

Suppose $$Ax=b$$, where $$A$$ is of dimension $$q\times p$$ with $$q, $$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!

• 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 $A$?) – littleO Aug 17 at 20:24
• Ah, I see it now. This is very helpful. Thank you so much! – TD888 Aug 17 at 23:49