When does $A^TAX = A^TB$? The original question is a true-or-false question: Assume $A$ is a $m\times n$ matrix and $B$ is a $m\times p$ matrix. If $X$ is an $n\times p$ unkwown matrix, then the system $A^TAX = A^TB$ always has a solution.
To my knowledge, for $A^TAX$ to always have a solution, $A^TA$ must be invertible. For it to be invertible, $A$ must have full column rank. However, it is not specified in the question, so I would assume it is false. However, the answer is true. Am I not getting something here? Thanks for any help!
Edit: This is the suggested answer given in the question:
Let $X$ = [$X_1$ $X_2$... $X_p$], $B$ = [$b_1$ $b_2$... $b_p$]
$A^TAX = A^TB$ $\Leftarrow$$\Rightarrow$ [$A^TX_1$ $A^TX_2$... $A^TX_p$] = [$A^Tb_1$ $A^Tb_2$... $A^Tb_p$]
Because normal equation $A^TAX_i = A^Tb_i$ must have a solution for all i = 1, 2, ..., p
So $A^TAX = A^TB$ must have a solution

We consider the case $p=1$ as below. 
Let $A$ be an $m\times n$ matrix and let $b\in\mathbb{R}^m$, and define 
$W=\{Ax:x\in\mathbb{R}^n\}$. 
Then it is well-known that there exists an unique vector in $W$, say $Ax_0$ where $x_0\in\mathbb{R}^n$, such that $b-Ax_0\in W^\perp$. Thus
$$0=\left\langle Ax,Ax_0-b\right\rangle
=\left\langle x,A^T(Ax_0-b)\right\rangle
=\left\langle x,A^TAx_0-A^Tb\right\rangle, \tag{1}$$
for all $x\in\mathbb{R}^n$. If taking $x=A^TAx_0-A^Tb$ into equation $(1)$,
we can get 
$$A^TAx_0-A^Tb={\it 0}\quad\mbox{or}\quad A^TAx_0=A^Tb.$$
Therefore the system $A^TAx=A^Tb$ always has a solution, and this can be easily extended to the case $p>1$.
