If $A$ and $B$ are $n^{\text{th}}$ order matrices and $\text{rank}$ of $AB$ is $1$ then $\text{rank}$ of $BA$ can't be $n$. Explain

Question

If $A$ and $B$ are $n^{\text{th}}$ order square matrices.

I think statement is true but I didn't know a clear reason and proof of it please explain. Got it as rank of AB <= mini(rank(A), rank(B)).....(1) rank(BA) <= min(rank(A),rank(B))......(2) also rank(A) +rank(B)-n<= rank(BA) or rank(AB)....(3) if rank of AB is 1 and if I let rank(A)=n then rank(B)=1 by (3) so by (2) rank(BA) is 1 if I let rank(A)=n-1 then rank(B)=2 and rank(BA)=2 {if n-1>2} if proceed in this manner then we observe that rank(BA) never come out to be n Because if it is n then by(2) both A and B have rank n and hence by (1) rank of AB must also be n which is contaradiction to given condition . Hope anyone who see my explanation understood it.

Answer 1

$\rho(AB)=n$ implies that $\rho(A)=\rho(B)=n$, which in turn implies $\rho(BA)=n$. Hence $n=1$.

Answer 2

Hint: $\DeclareMathOperator{\rk}{rank}$

We have the general inequality $$\rk(BA)\le \rk A,\;\rk B.$$ What happens if $\rk(BA)=n$?