Product of matrices of different order is not invertible If $A$ is an $m \times n$ and $B$ is an $n \times m$ matrix with $n < m$ then prove that $C = AB$ is not invertible.
I am not getting any idea. May be I am missing vary basic point. $C = AB$ where both $A$ and $B$ are square matrices  will be invertible iff both $A$ and $B$ be invertible. I know it and its proof.
May be this question has been discussed earlier, but I am not seeing the link. Thank you for your help.
 A: Hints: The rank of a $j\times k$ matrix is at most $\min\{j,k\}$. If the ranks of matrices $M,N$ are $s,t$ (respectively) and they are sized appropriately for $MN$ to make sense, then the rank of $MN$ is at most $\min\{s,t\}.$
A: Hint: Suppose $C$ is an inverse of $AB$. Consider what the linear transformation $v\mapsto CAv$ does to the vectors $B\mathbf e_1,B\mathbf e_2,\ldots,B\mathbf e_m\in\mathbb R^n$. Are these vectors linearly dependent? Are their image after $CA$? Is it possible for a linear transformation to map a linearly dependent family of vectors to a linearly independent one?
A: One way to see this is that if a composite function $f=g\circ h:X\to Y$ is to be bijective, it is necessary (though not sufficient) that $h$ be injective (if you would start mapping two elements of$~X$ to the same image, no further map could make their images distinct again and $f$ would not be injective), and also that $g$ be surjective (if some $y\in Y$ would fail to be in the image of $g$ it would also fail to be be in the image of$~f$, and $f$ would not be surjective). Now if $n<m$, you know that (the linear map corresponding to) a $n\times m$ matrix like $B$ cannot be injective, and also that a $m\times n$ matrix like $A$ cannot be surjective, which gives you two reasons why the composition $C=AB$ cannot be bijective (i.e., invertible).
A: More hints in the form of words:
When you multiply two matrices, the rank can't increase.
Rank can't be more than the smaller of the number of rows and number of columns
For an invertible matrix, the rank should equal the size
