There do not exist $m\times n$ and $n\times m$ rectangular matrices $A$ and $B$ . t. $AB=I_m$ (The $m\times m$ Identity matrix) 
Let $m,n\in\Bbb N$ be positive with $m>n$. Prove that $\nexists A\in M_{mn}\ \&\ B\in M_{nm},\ AB=I_m$ (The $m\times m$
  Identity matrix)

I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=\begin{bmatrix}
1 &0  &0 \\ 
 0&  1& 0
\end{bmatrix}$ and $A=\begin{bmatrix}
1 &0 \\ 0&1 \\ 0&0 \end{bmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradict with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So, a proof for the given question is highly appreciated 

Let $B$ be a matrix with more columns $B_1,\ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = \sum_{i \not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k \cdot B_k = 1$ and $A_k \cdot B_i = 0$ for each $i \not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_l\cdot B_l = 1$ and $A_l \cdot B_i = 0$ for all $i \not = l$. However, $A_l \cdot B_i = 0$ for all $i \not = l$ would imply $A_l \cdot B_l = \sum_{i \not = l} c_iA_l \cdot B_i = 0$, contradicting the equation $A_l \cdot B_l = 1$.
