I've found proofs for this claim but they were too complicated so I came up with my own, though I don't know if it's enogh, would like to get feedback. thank you.

So there is claim saying that

We have a linear system and a matrix A in the order m x n

There is only one solution for Ax=b if and only if rankA=n (n the number of unknowns)

Proof;

(<=) So we know that the dimension space of the solution for Ax=0 is n-rankA (proved it before so I am relying on this fact)

so it is given that rankA=n so we get the dimension of the solution for homogenous system 0, in other words the only solution for Ax=0 is when x=0

And we know that the solution for the non homogenous system Ax=b is in the form x+u when u is the solution for the homogenous system, but we showed that the only solution for homogenous system is 0 so our only solution for Ax=b is x (x+u=x+0=x)

(=>)So if the linear system has only one solution we have to prove that n=rankA

We will assume that rankA<n

So if rankA<n then we get that n-rankA>0 and we know that n-rankA is the dimension of the solution space of the homogenous system and it's bigger than 0 (According to the assumption)

So we know the solution for Ax=b is in the form x+u when u is the solution for homogenous system. And we showed that there are more vectors than zero vector in the homogenous solution space, so lets take u1, u2 that are not equal to each other A(x+u1)=Ax+Au1=b+0=b and A(x+u2)=Ax+Au2=b+0=b

So we got two different solutions and it's a contraddiction to being only solution and therefore RankA=n

So is this correct proof? Thanks