# Product of two elements is identity implies they are mutual inverses..

Let $A$ be an associative unital n-dimensional algebra over field $F$.

Show that if $ab=1$ for some $a,b \in A$ then $a=b^{-1}$

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$L$ is surjective, in fact a linear operator on the vector space $A$ of dimension $n$ is surjective iff it's injective, which is a consequence of the dimention formula $dim(im(L))+dim(Ker(L))=n$ – user31899 Aug 11 '12 at 7:16
OP may be talking about $\mathrm{End}_A(A)\cong A^{op}$, not $\mathrm{End}_k(A)$ (because of the comment about considering $A$ as a left module over itself). What confuses me is that $\mathrm{End}_A(A)=A^{op}$ (because $a\mapsto\phi(a)=a\phi(1_A)$), and $A^{op}$ is not always isomorphic to $A$, so I wonder what $A\to\mathrm{End}_A(A)$ is supposed to be. Otherwise $\mathrm{End}_k(A)\cong M_n(k)$ has $\dim=n^2$ when $\dim_k A=n$, so $A\to\mathrm{End}_k(A)$ is certainly not surjective. – anon Aug 11 '12 at 7:40

We consider the map $f:A \rightarrow A$ sending $x$ to $bx$. $f$ is clearly $F$-linear. If $f(x) = 0$, $af(x) = abx = 0$. On the other hand, since $ab = 1, abx = x$. Hence $x = 0$. This means that $f$ is injective. Since $A$ is finite dimensional over $F$, $f$ is surjective. Hence there exists $y \in A$ such that $by = 1$. Hence $ba = ba(by) = b(ab)y = by = 1$. Therefore $a = b^{-1}$.
idea: Call the representation of $A$ by $\phi$, then $ab=1$ ,since $\phi(ab)=id_A$ and $A$ is finite dimensional implies $\phi(ba)=id_A$
This new proof is probably not fine. It requires a faithful, finite-dimensional representation $\phi$. Your old proof specifically addressed both points. – Jack Schmidt Aug 11 '12 at 21:08