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.
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Assuming $A$ is invertible... – Jun 01 '14 at 10:24
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2See here. – Ragib Zaman Jun 01 '14 at 10:26
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1This is possible, provided $A$ and $B$ are not square. – egreg Jun 01 '14 at 10:28
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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 $BA\ne I_2$.
egreg
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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 $$\pmatrix{1&0\cr0&0\cr0&1\cr}\pmatrix{1&0&0\cr0&0&1\cr} =\pmatrix{1&0&0\cr0&0&0\cr0&0&1\cr}\ .$$
David
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