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

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 $$\operatorname{rank}(A) we can get no solution for the equation.
What I wanted to ask if the given information was for every $$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.

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$$\dim(\ker A)=\operatorname{rank}A=n\iff 0+m=n.$$