Let $A$ be $m\times n$ and $B$ be $n\times m$ matrix over real number with $m
Let $A$ be $m\times n$ and $B$ be $n\times m$ matrix over real number
  with $m<n$. Then
  
  
*
  
*$AB$ is always nonsingular.
  
*$AB$ is always singular.
  
*$BA$ is always nonsingular
  
*$BA$ is always singular.
  



*

*False, Since $A=0$, $B=0$ $\implies$ $AB=0$.

*Need not be true. Let $x$ be $1\times n$ matrix. Since, $x x^T\ge 0$

*Using the same reasoning 3 can be false.
What is the guarantee that $BA$ is always singular? If $A=x\neq 0, B=x^t.$ Then $BA$ is singular. How do I prove In general?

$BA \in \mathbb{R}^{n \times n}$,
but $rank(BA)\le \min\{rank(A) ,rank(B) \} \le m<n$
Hence it must be singular.
