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Let $V$ be a vector space and $A,B\in\mathcal L(V)$, the set of linear operators on $V$. Suppose $AB=I$. Must $BA=I$? In other words, do left and right inverses coincide? If not, do conditions of finite-dimensionality of $V$ or boundedness of $A$ and $B$ change anything?

I am asking because I know that for a group, the existence of left inverses or right inverses alone guarantees existence of the other, with both coinciding. But the set of operators with, say, left inverses, doesn't automatically form a group, so I am not sure whether this applies. But I can't think of any example of square matrices with different left and right inverses.

Arturo Magidin
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WillG
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    “In a group” is the wrong context. In a semigroup, if you have left inverses and left identity, then you have two-sided inverses; if you have right inverses and a right identity, then you have two-sided inverses. But if you are in a group, then everything automatically has two-sided inverses. – Arturo Magidin Feb 12 '21 at 14:52
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    See e.g. here for counterexamples. It is true if the vector space is finite dimensional. – Robert Israel Feb 12 '21 at 15:18

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For infinite dimensional vector spaces it may fail. A standard example is to take the space of real sequences, let $B$ be the right shift operator, and $A$ the left shift operator, $$\begin{align*} B(x_1,x_2,x_3,\ldots) &= (0,x_1,x_2,x_3,\ldots)\\ A(y_1,y_2,y_3,\ldots) &= (y_2,y_3,y_4,\ldots) \end{align*}$$ For finite dimensional vector spaces, $AB=I$ imlies $BA=I$: $B$ is one-to-one, hence surjective, hence invertible.

Arturo Magidin
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