product of m x n matrix with n x m matrix How to prove that product of  $\mathbb{m x n}$ matrix with $\mathbb{n x m}$ matrix is not invertible given $\mathbb{m >n}$. 
For the case of $\mathbb{2 x 1}$ and $\mathbb{1 x 2}$ matrix, it is clear; since for the product matrix A; $\mathbb{AX=0}$ has non trivial solutions. (both the resulting equations turn out to be same after cancellation of common factors.
How to go about proving it for the general case?
 A: Let $A$ be $m\times n$, and let $B$ be $n\times m$. Let $C=AB$. $C$ is $m\times m$, so it’s invertible if and only if its rank is $m$, i.e., if and only if the map
$$T_C:\Bbb R^m\to\Bbb R^m:x\mapsto Cx$$ 
is a surjection. (In possibly more elementary terms that’s saying that the column space of $C$ is all of $\Bbb R^m$.) But $T_C=T_A\circ T_B$, where
$$T_A:\Bbb R^n\to\Bbb R^m:x\mapsto Ax$$
and $$T_B:\Bbb R^m\to\Bbb R^n:x\mapsto Bx\;.$$ That is, $T_C$ can be decomposed as
$$\Bbb R^m\overset{T_B}\longrightarrow\Bbb R^n\overset{T_A}\longrightarrow\Bbb R^m\;.$$
If $T_C$ is a surjection, $T_A$ must also be a surjection, and that’s impossible: the range of $T_A$ is the column space of $A$ and therefore has dimension at most $n$, and $n<m$.
A: For any given matrix, the dimension of it's row vectors is equal to the dimension of its column vectors. Therefore, since m>n, both matrices have rank at most 'n'.
The rank of a product of two matrices can be no bigger than the minimum rank of either matrices, since every matrix multiplication is a linear combination of rows/columns of either matrix. So the product is of rank at most 'n' but an m*m matrix is invertible iff it is of rank m
