Linear system with non-square matrix

Question

I have a matrix $A$ $m \times n$, and $\forall \ b$ the linear system $Ax = b$ has a $\mathbf{unique}$ solution. I need to prove that $\forall \ c$ linear system $A^Tx = c$ has also a unique solution.

$\mathbf{My \ proof\ is}$: Since $Ax$ ($x = (x_1 x_2 \dots x_n)^T$) is the linear combination of columns of the matrix $A$ with coefficients $x_1, x_2, \dots \ $and $\forall \ b \in \mathbb{R^m}$ we only have a unique set of coefficients $x_1, x_2, \dots \ $of the linear combination hence columns of $A$ form a basis in $\mathbb{R^m}$, consequently $A$ is the square matrix $\rightarrow \ A^T$ is the square too. Also $rankA = m \rightarrow det(A) \neq 0 \rightarrow det(A^T) \neq 0 \rightarrow \forall \ c$ system $A^Tx = c$ has a unique solution.

Could you please tell me whether this is a correct proof? Thank in advance!

Answer 1

Your proof ultimately amounts to the following:

• Because $Ax = b$ has a unique solution for all $b$, $A$ is square with full rank.
• Because $A$ is square with full rank, $A^T$ is square with full rank.
• Because $A^T$ is square with full rank, $A^Tx = c$ has a unique solution for all $c$

This is a valid line of reasoning.