Question about a statement involving system of linear equations and rank of matrix.

Question

the statement is:

let $A\in \mathbb{R}^{m\times n}$, if for every $c\in \mathbb{R}^m$ there exists a solution for $Ax=c$, then $\operatorname{rank}(A)=m$.

Now I can understand that this is a true statement, since if $\operatorname{rank}(A)<m$ we can get no solution for the equation.
What I wanted to ask if the given information was for every $c\in \mathbb{R}^m$ there exists a unique solution, could I infer that the matrix is square? ($m = n$).
This is coming from the idea that if $Ax=c\:$ has a unique solution, then $\operatorname{rank}(A) = n$, so if it has a solution for every $c$ that means $\operatorname{rank}(A) = m $ and so $m=n$.
I would appreciate any feedback and would love to know if there's more interesting stuff to infer from the given information.

Answer 1

Yes, it is true. Note that the rank of the matrix $A$ is also the dimension of the subspace generated by the columns of the matrix, and if evety vector in $\mathbf R^m$ is attained, this dimension is $m$.

On another hand $\operatorname{rank}A\le \min(m,n)$, and if the solution is unique; thr rank-nullity formula shows thatt the rank is the codimension of the kernel of the associated linear map, i.e. ˆ$$\dim(\ker A)=\operatorname{rank}A=n\iff 0+m=n.$$

Answer 2

If the rank m>n, the system of equations is overdetermined, there are too many variables relative to the number of equations, and there are no solutions.

If the rank m<n, the system of equations is underdetermined (there are too few non-zero variables compared to the number of equations), and there are an infinite number of solutions.

There is a unique solution iff the rank m=n.