Why is only a square matrix invertible? Can anyone give a very simple proof (or explanation) as to why only square matrix can possibly be invertible?
 A: A product of two matrices of order $m\times n$ and $n\times p$ is a matrix $m\times p$. 
But when a matrix $A$ has an inverse $B$, it has a two sided inverse, that is $AB=BA=I$. 
The only possibility is $m=n=p$.
A: Basically, an $\,n\times m\,$ matrix represents a linear map between linear spaces over some field of dimensions $\,m\,,\,n\,$ .
That a matrix is invertible means the map it represents is invertible, which means it is an isomorphism between linear spaces, and we know this is possible iff the linear spaces' dimensions are the same, and from here $\,n=m\,$ and the matrix is a square one.
A: An inverse of a square matrix $A$ is $B$ such that $A B = I$. You can also find a $m \times n$ matrix $A$ and $n \times m$ matrix $B$ such that $A B = I$, and call $B$ inverse of $A$. However such inverse need not be unique, nor does it endow any subset of matrices with a group structure under multiplication. Therefore it isn't as interesting.
A: An $m\times n$ matrix $A$ is invertible if there exists an $n\times m$ matrix $B$ such that the products $AB=I_m$ and $BA=I_n$ are identity matrices. I'll show that this implies $m=n$. I'll work over an arbitrary commutative ring with $1\neq0$ (e.g. fields, the integers, all integral domains, etc.).
I just need the following basic properties of the determinant of square matrices:

*

*$\det(CD)=\det(C)\det(D)$ for any two square matrices $C$ and $D$.

*$\det(I)=1$ for $I$ the identity matrix of any size.

*$\det(C)=0$ if $C$ has any zero row or column.

We argue by contradiction. Assume that $m>n$ (one can do assume without loss of generality, since the roles of $A$ and $B$ can be exchanged). We extend $A$ and $B$ to $m\times m$ square matrices by adding zeroes,
$$C=(A|0),\qquad D=\left(\frac{B}{0}\right).$$
Then
$$CB=AB=I.$$
By 1, 2, 3, above
$$1=\det(I)=\det(CD)=\det(C)\det(D)=0,$$
reaching the desired contradiction.
