TITLE: Prove that if $ab = 1$ for $a, b \in A$, then $a = b^{-1}$.
QUESTION [2 upvotes]: Let $A$ be a unital finite-dimensional algebra over a field $\mathbb{K}$ and $a,b \in A$. I have a question to the following statement:

If $ab = 1$, then $a = b^{-1}$.

Originally, I thought this follows from definition, but I guess we have to show uniqueness? I asked my lecturer for some advice and he told me to think about how to represent $A$ in matrices. I am lost. Any pointers on how to approach this, or perhaps how to break this down so I can accumulate an understanding of how this works? Thanks heaps.

REPLY [0 votes]: Consider the $\mathbb{K}$-linear map $\lambda_b\colon A\to A$, $\lambda_b(x)=bx$. Similarly you have $\lambda_a$ and, by assumption,
$$
\lambda_a\circ\lambda_b=\iota
$$
where $\iota$ is the identity on $A$. Therefore $\lambda_b$ is injective. Since $A$ is finite dimensional, $\lambda_b$ is also surjective. Therefore $\lambda_b$ has an inverse $\mathbb{K}$-linear map $f$ and we need to prove $f=\lambda_a$.
This is easy:
$$
\lambda_a=\lambda_a\circ\iota=\lambda_a\circ(\lambda_b\circ f)=
(\lambda_a\circ\lambda_b)\circ f=\iota\circ f=f
$$
No matrices needed.
{"set_name": "stack_exchange", "score": 2, "question_id": 2937172}