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Let $A$ be a finite dimensional unital algebra over a field $K$. Let $a,b \in A$ such that $ab=1$, show that $ba=1$

One way to prove this is by defining multiplication map and then using Rank nuility theorem. Is there any other way using matrices or something else?

Dontknowanything
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  • I suspect that, choosing a vector space basis of $A$, you can construct an isomorphism onto some algebra of matrices, and then the property follows from the analogous property of matrices. – Giuseppe Negro Sep 24 '18 at 09:13
  • Note that the proof suggested by Giuseppe Negro is essentially the same as the one given in the question, so matrices don't really give any new insight. – Arnaud D. Sep 24 '18 at 09:43
  • The duplicate I picked has at least two answers that address exactly this. There are more specific duplicates like this one but I feel their content is already subsumed in the duplicate I chose. I found them by searching "finite dimensional ab=1" – rschwieb Sep 24 '18 at 10:49

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Fix a vector space basis $\{a_1, a_2, \ldots, a_n\}$ of the algebra $\mathfrak A$. Then to every $a\in A$ it corresponds a unique $n\times n$ matrix $A$ by the formula $$ \sum_{ij}(A^i_j x^j )a_i = a(\sum_h x^h a_h).$$ To the product $ab$ there corresponds the matrix product $AB$. So this map is an injective algebra homomorphism, hence it is an isomorphism of $\mathfrak A$ onto an algebra of matrices. The property about left and right inverses in $\mathfrak A$ now follows from the analogous one for matrices.

Giuseppe Negro
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  • Oh sorry, I noticed now that there is a notational conflict: the letter $A$ denotes both the algebra and the matrix corresponding to $a$. I am changing the algebra's name to $\mathfrak A$. – Giuseppe Negro Sep 27 '18 at 08:05