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 
 A: Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
A: The column space of an $m\times n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $\mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
\{Ax:x\in\mathbb{R}^n\}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $AB\ne I_m$.

The dimension of the column space of $A$ is commonly known as the rank of $A$, $\operatorname{rk}(A)$. The above argument can then be stated as $\operatorname{rk}(AB)\le\operatorname{rk}(A)$. It also holds that $\operatorname{rk}(AB)\le\operatorname{rk}(B)$ (not needed for the above proof).
A: 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 $A_l \cdot B_l = 1$.
