Matrices whose product is identity but do not commute.

Question

I'm supposed find two matrices $A$ and $B$ whose product $AB=I_2$, but $BA\neq I$. But I'm not sure if this is even possible since if $AB=I$, doesn't that mean that $B$ is the inverse matrix of $A$ and that leads to $BA=I$ automatically. Any help is appreciated.

This is possible, provided $A$ and $B$ are not square. – egreg Jun 01 '14 at 10:28 — egreg, Jun 01 '14 at 10:28

score 6 · Accepted Answer · answered Jun 01 '14 at 10:30

Saying that $AB=I_2$ means that $A$ has two rows and its rank is $2$ (it's a necessary and sufficient condition for the existence of a right inverse). If $A$ is square, then $AB=I_2$ implies $BA=I_2$. If $A$ is non square, then $AB=I_2$ implies $BA\ne I_2$.

score 3 · Answer 2 · answered Jun 01 '14 at 13:24

This is not possible if $A$ and $B$ are $2\times2$ matrices. However, $$\pmatrix{1&0&0\cr0&0&1\cr}\pmatrix{1&0\cr0&0\cr0&1\cr}=\pmatrix{1&0\cr0&1\cr}$$ while $$\pmatrix{1&0\cr0&0\cr0&1\cr}\pmatrix{1&0&0\cr0&0&1\cr} =\pmatrix{1&0&0\cr0&0&0\cr0&0&1\cr}\ .$$

Matrices whose product is identity but do not commute.

2 Answers2