I was trying to figure out the following at work today.
Suppose $\tau\in\mathscr{L}(V)$, for $V$ an $n$-dimensional vector space over a field $F$. Let $A\in M_n(F)$, and $\sigma\colon V\to F^n$ be an isomorphism such that $\sigma\tau=A\sigma$. Why does there exist an ordered basis $\mathcal{B}$ such that $A=[\tau]_\mathcal{B}$?
I rewrite $\sigma\tau=A\sigma$ as $\sigma\tau\sigma^{-1}=A$. I know the $i$th column of $A$ is given by $ A^{(i)}=Ae_i=\sigma\tau\sigma^{-1}(e_i). $ Moreover, supposing $\mathcal{B}$ exists, $[\tau]_\mathcal{B}=([\tau b_1]_\mathcal{B}|\cdots|[\tau b_n]_\mathcal{B})$, so I should have $ [\tau b_i]_\mathcal{B}=\sigma\tau\sigma^{-1}(e_i). $ However, I don't know how to choose $\mathcal{B}$ so that the vectors $b_i$ satisfy this. How could one find such a basis $\mathcal{B}$? Thank you.