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I am trying to prove that for two linear transformations ($A$ and $B$), if $AB + I$, then $BA = I$ as well. I think that the best way to go about this is by using the fact that all linear transformations can be represented with a matrix, then show that $AB = BA = I$ whenever $B = A^{-1}$ (note I am working with finite vectors, I am not sure of the properties in an infinite case).

How can you prove that $AA^{-1} = I$ given that $A^{-1}A = I$?

Arthur
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  • Surprisingly, I wasn't able to find this via searches. I thought it would be simply but I don't see how to go about this with n vectors. –  Feb 17 '16 at 12:30
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    If you know that there are $B, C$ such that $AB = CA = I$, then it's not very difficult to show that $B = C$. But showing the existence of $B$ by using the existence of $C$ is not trivial. – Arthur Feb 17 '16 at 12:39

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Think of the matrices as linear maps:

If $f \circ g = \operatorname{id}$, then $g$ is well-known to be injective. On a finite-dimensional vector space, we deduce that $g$ is also bijective, i.e. there is some map $h$ such that $g \circ h=\operatorname{id}$. Now we have

$$h = (f \circ g) \circ h = f \circ (g \circ h)=f,$$

which is the desired result.

Note that we have not used that $h$ is also linear. It can be any map, but a posteriori we get that $h$ is indeed linear, since we proved $h=f$.

MooS
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