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In Serge Lang's Linear Algebra, page 35 it states: Let $A$ be a square $n\times n$ matrix. We shall say that $A$ is invertible or non-singular if there exists an $n\times n$ matrix $B$ such that $BA=AB=I_{n}$. -- definition (i)

But, I think in the definition above, some information provided is redundant, because if we know $BA=I_{n}$, we can prove $AB=I_{n}$. As a result, we needn't assume $AB=I_{n}$ and the definition should be modified as following:

Let $A$ be a square $n\times n$ matrix. We shall say that $A$ is invertible if there exists an $n\times n$ matrix $B$ such that $BA=I_{n}$. -- definition (ii)

However, almost all the textbooks use definition (i) rather than definition (ii) and I don't know why. I'm so confused. Could you help me to understand why textbooks use definition (i) rather than definition (ii) ?

studyhard
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    The first definition is the correct definition of invertibility in general, e.g. you need it for infinite-dimensional vector spaces. It's a nontrivial and in some sense misleading theorem that in the finite-dimensional case the second definition suffices. – Qiaochu Yuan Aug 15 '24 at 06:25
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    Also, the proof of the fact that $BA=I_n$ implies $AB=I_n$ may come much later in the books. It is not immediate from the definition. – Kavi Rama Murthy Aug 15 '24 at 06:27
  • Maybe not a duplicate, but probably of interest: https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – Andrew D. Hwang Aug 15 '24 at 13:03

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Lang is ruling out a left- inverse or right-inverse. In other words, if $BA=I_n$, it is possible that B is an $n\times m$ while A is $m \times n$ (both are rectangular, not full-rank matrices), in which case, $AB \neq I_n$. Only when $AB=I_n$ as well, do we then say that A and B are each others inverse.

user1140514
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