Are there matrices such that $AB=I$ and $BA \neq I$ Are there matrices such that $AB=I$ and $BA \neq I$ ? 
$A$ and $B$ are square matrices 
 A: With square matrices, no. With non-square matrices, it's perfectly possible. For example,
\begin{align*}
A &= \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right) \\
B &= \left(\begin{array}{cc}1 & 0 \\ 0 & 1 \\ 0 & 0\end{array}\right)
\end{align*}
Note that, indeed, we cannot have non-square matrices where $AB$ and $BA$ are identities of appropriate dimensions, because that would imply the existence of an isomorphism between spaces of different dimensions!
A: If $AB=I$ with $A,B$ square matrices, then $\ker(B)=\{0\}$. Therefore $B$ is also surjective by rank-nullity, so $B$ is invertible and $A=B^{-1}$, so $BA=I$.
A: If $A$ and $B$ are finite square matrices (of size $n\times n$), then $$AB=I_n\implies BA=I_n$$
However, if $A$ and $B$ are not square, if $AB=I_n$, then $BA\neq I_n$ in general (and $BA$ need not even be the same size as $AB$).
This link may provide more information on invertible matrices.
A: Answer to your question: Assuming $A,B$ are squares, the answer is there are none.
Claim: Let $\mathbb F$ be any field. If $A,B\in \mathbb F^{n\times n}$, then $AB=I$ implies $BA=I$.
Proof:
We first show that $A$ is nonsingular. Suppose otherwise. Then there exist elementary matrices $E_1,\ldots, E_k$ such that 
$$E_k\cdots E_1A=C$$
where $C$ has a zero row. Then
$$E_k\cdots E_1AB=CB$$
has a zero row. The matrix $AB=I$ is row equivalent to $CB$, which is absurd since $I$ has no zero row. Thus, $A$ is nonsingular.
Since $A$ is nonsingular,
$$B=(A^{-1}A)B=A^{-1}(AB)=A^{-1}\cdot I=A^{-1}.$$
Hence, $BA=I$ too.
