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A thread I saw recently has led me to believe that this is not a valid proof of the fact that for matrices $A$ and $B$, $AB=I\implies BA=I$.

Suppose $AB=I$. Then

$$A^{-1}AB=A^{-1}I$$

$$B=A^{-1}$$

$$BA=A^{-1}A$$

$$BA=I$$

what step is wrong in this? I assume $A$ has an inverse because $\det A\det B=\det AB=\det I=1$, so $\det A\neq 0$.

Alex Mathers
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  • What leads you to believe this is invalid? I've seen exactly this proof in Schaum's Linear Algebra text. – Dion Bridger Jun 27 '15 at 23:43
  • http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i It is correct since the existence of a left inverse of a square matrix follows from the existence of a right inverse. – Hasan Saad Jun 27 '15 at 23:43
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    How do you know that $A$ has an inverse? Once you do, then of course $B$ is it. But the problem is to start with $AB=I$, not knowing that $A$ has an inverse, and conclude $BA=I$. And from this it will follow that $B$ is the inverse of $A$. – GEdgar Jun 27 '15 at 23:43
  • Are you assuming that $A$, $B$, and $I$ are all $n\times n$ matrices? If that is not assumed, your statement is false and the proof is invalid since some of the products will be undefined. Is that what you mean? – Rory Daulton Jun 27 '15 at 23:43
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    Pretty sure he's assuming they're square. Else, this would have been obviously faulty to anyone. – Hasan Saad Jun 27 '15 at 23:44
  • The question I think you might be thinking about ([this](http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i)) asks for proofs where there are some restrictions imposed (i.e. if you don't know about inverses). – Winther Jun 27 '15 at 23:45
  • @HasanSaad that's what's being proved here though? And to the others, yes I am assuming they're square and same dimension. – Alex Mathers Jun 28 '15 at 00:31
  • Also, thanks @Winther that helps – Alex Mathers Jun 28 '15 at 00:32
  • The first equality may be incorrect. There is no assumption that $A^{-1}$ exists – sleeve chen Jun 28 '15 at 00:33
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    @sleevechen For a square matrix, the existence of a right inverse implies the existence of a left one. As for OP, the proof here is that if $AB=I$ then $BA=I$ and this proof is completely valid. However, this "proof" misses the gist of the subject which is all about proving the existence of $A^{-1}$ first. If it were proved before, then this proof is perfectly valid. – Hasan Saad Jun 28 '15 at 00:47

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If $A$ and $B$ are both square matrices of the same dimension, then your proof is certainly correct. However, $AB=I\not\Longrightarrow BA=I$ when $B$ and $A$ are not square, and your step of $\det(AB)=\det(A)\det(B)$ is wrong since $\det(A)$ and $\det(B)$ are not defined for non-square matrices. In general for a non-square matrix there are so-called left- and right-inverses, which may not be identical for a given matrix.

For example, let $$A=\begin{pmatrix}1&1&0\\0&0&1\end{pmatrix}, B=\begin{pmatrix}1&0\\0&0\\0&1\end{pmatrix}$$ You can check $AB=I_\text{2x2}$ but $BA\neq I_\text{3x3}$

Leo
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