# If $A\sigma=\sigma\tau$, does $[\tau]_B=A$ for some basis $B$?

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.

• Are you studying Steven Roman's book "Advanced Linear Algebra"? Dec 13 '12 at 4:24
• @Manos Yeah, this is problem 2.7 I think. Dec 13 '12 at 4:36
• I could tell from the notation. Awesome book :) Dec 13 '12 at 4:37
• @Manos I agree. For the most part, his presentation is impeccably clear. Thanks for your answer! Dec 13 '12 at 4:47
• You are welcome :) Dec 13 '12 at 4:48

Since $\sigma: V \rightarrow F^n$ is an isomorphism, let $b_i=\sigma^{-1}(e_i)$, where $e_i$ is the standard $i^{th}$ basis vector of $F^n$. Define $B=\left\{\sigma^{-1}(e_1), \cdots, \sigma^{-1}(e_n)\right\}$. Then for any vector $v \in V$ we have by construction that $[v]_B=\sigma(v)$. Now $[\tau b_i]_B = [\tau \sigma^{-1}(e_i)]_B=\sigma \tau \sigma^{-1}(e_i)=A \sigma \sigma^{-1}(e_i)=A e_i = A^{(i)}$.