Linear algebra, construction of counter example for two rectangular matrices such that $AB=I$ but $BA\neq I$

Question

We have a matrix $A$ of order $m$ by $n$ and a matrix $B$ of order $n$ by $m$ where entries are from real numbers. We are given that $AB =I$ and we need to check whether $BA =I$ or not. If we have square matrices then it is true but here are rectangular matrices. I know that it doesn't hold true for rectangular matrices and I am trying hard to find some counter example but unable to find such matrix. I want to learn about how to construct matrices as counter examples in linear algebra as in this case. Thanks

Answer 1

Take $A=\begin{pmatrix}1&0\end{pmatrix}$ and $B=\begin{pmatrix}1\\0\end{pmatrix}$. Then$$A.B=\begin{pmatrix}1\end{pmatrix}\text{ and }B.A=\begin{pmatrix}1&0\\0&0\end{pmatrix}.$$

Answer 2

Take $A=(1,0)$ and $B=A^T$.

Compute $AB$ and $BA$.

Answer 3

Let $A$ be an $n\times m$ matrix, and $B$ is $m\times n$, with $n\ne m$. If $AB=I_n$ (which is a $n\times n$ matrix), then $rank (AB) = n$, hence $rank(A)= n$ and $rank(B)= n$. This implies that $n<m$.

Now, the matrix $BA$ is a $m\times m$ matrix, but since $rank(B)=n<m$, you have necessarily $rank(BA)<m =rank(I_m)$, so $BA$ can never be equal to $I$.