Proof check: AB not invertible

Question

Let $A$ be an $n\times m$ matrix and $B$ an $m\times n$ matrix, both with entries in a field $F$. If $m<n$, show that $AB$ is not invertible.

My attempt: Suppose on the contrary that $AB$ is invertible, and consider the rank-nullity theorem. $\operatorname{Rank} AB + \operatorname{Null} AB=n$, and since $AB$ is invertible, $\operatorname{Null} AB=0$, so $\operatorname{Rank} AB=n$. Then $\operatorname{Rank} A + \operatorname{Null} A=m<n=\operatorname{Rank} AB$. Let $v$ be in the range of $AB$, then $ABw=v=A(Bw)$ for some $w$, so $v$ is also in the range of $A$. Then $\operatorname{Rank}AB\leq\operatorname{Rank}A$. So now we have $\operatorname{Rank} A + \operatorname{Null} A<\operatorname{Rank} AB\leq\operatorname{Rank}A$, which implies that $\operatorname{Null}A=0$ and $\operatorname{Rank}A=\operatorname{Rank}AB$, a contradiction.

Answer 1

Since $m<n$, we have $rank(A)\le m$ and $rank(B) \le m$.

$AB$ is of dimension $n\times n$ and $rank(AB)\le \min\{ rank(A),rank(B)\}\le m<n$. It is not full rank hence it is not invertible.

Answer 2

Here is a solution not involving rank, based on elementary matrix knowledge :

We have $PA=R$ where $P$ is an invertible $n\times n$ matrix and $R$ is an $n\times m$ row-reduced echelon matrix. Since $n \gt m$, $R$ has at least one row of zero's, namely the $n^{th}$ row.

Suppose now that $AB$ is invertible, then $PAB$ is invertible as a product of invertible matrices. But $PAB=RB$ so the $n^{th}$ row of $PAB$ is a row of zero's, a contradiction.