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I am asked by StackExchange to clarify if this is a duplicate question. This is not a duplicate of the question about matrix multiplication. This question is not (on its surface) a question about matrices.

I am beginning to learn Geometric Algebra.

Let $A, B$ be multivectors in Geometric Algebra $\mathbb{G}^n$. $AB = 1$ iff $BA = 1$. Why?

Progress: Let $e_{I}$ be a canonical basis vector, and $|I|$ is odd. Let $P$ be the projection operator onto this basis vector. Then $P[AB] = P[BA]$. The case of $|I|$ even gets messy. I do know for any two multivectors, the scalar part $S$ has the property $S[AB] = S[BA]$

Mark
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    Possible duplicate of If $AB = I$ then $BA = I$ – rschwieb May 17 '17 at 03:07
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    It's important to remember geometric algebras are just extremely tame semisimple rings. There is nothing very special about them structurally. They're all either matrix algebra over. The reals complexes or quaternions, or the square of such a ring. What I mean is that quickly diving into specific proofs using their features is usually a big distraction. – rschwieb May 17 '17 at 03:09
  • Thank you @rschwieb. I have actually not heard of the term semi simple ring before and I will have to investigate more. Even after looking at Wikipedia I still don't completely understand, nor is it clear from the definitions that there is a connection to matrix multiplication. Nor is it clear how to answer my question from the linked candidate duplicate (at least without LordSharkTheUnknown's answer). I hope this question can remain not marked as duplicate. – Mark May 17 '17 at 08:11
  • All that really needs to be said is that your ring is a finite dimensional algebra, or even just Artinian, then all of the numerous solutions at the dupe handle your special case. It needs to be linked to he existing answers one way or another. Regards – rschwieb May 17 '17 at 10:40
  • @rschweib oh I was under the impression that this answer could be deleted because of duplication. If it's just a link I agree – Mark May 17 '17 at 14:38
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    In rare cases they could be deleted... It's certainly no automatic. if you look around you'll find quite a few examples of highly duplicated questions. Their duplicates are valuable in that they make it easier to uncover all sorts of answers that are already out there. – rschwieb May 17 '17 at 16:59

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A geometric algebra (these are basically the same as Clifford algebras) is a finite-dimensional algebra over a field. The implication $ab=1$ implies $ba=1$ always holds in such an algebra. One way to see this is that all such algebras have faithful representations as algebras of matrices, and $AB=I$ implies $BA=I$ is well-known for matrices.

Angina Seng
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  • Can you elaborate on the matrix representation? I imagine it would be completely specified by the matrices of the canonical basis vectors. What do those matrices look like? Thanks! – Mark May 14 '17 at 18:50
  • Your question seems to exclude matrices which only a left or right inverse. Does your answer imply the matrices are square? – dantopa May 14 '17 at 18:52
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    @dantopa I speak of algebras of matrices. Algebras involve addition and multiplication. To add two matrices they must have the same shape: so an algebra of matrices consists of matrices of some fixed shape. To multiply matrices of the same shape, they must be square. – Angina Seng May 14 '17 at 18:54
  • @LordSharktheUnknown I have posted my comment as a separate question. https://math.stackexchange.com/questions/2281006/representation-theorem-for-finite-dimensional-algebra-over-fields?noredirect=1&lq=1

    When that is addressed I will accept this answer. Thanks!

     – Mark May 14 '17 at 19:39