Are there matrices such that $AB=I$ and $BA \neq I$ [duplicate]

Solution 1:

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!

Solution 2:

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 $BA=I$.

Solution 3:

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.