Show that $[U]_\beta={[T]_\beta}^*$ for any orthonormal basis $\beta$

Question

Let $V=F^n$ and let $A \in M_{n*n}(F)$ and $A^*$ mean conjugate transpose.
Define linear operators $T$ and $U$ on $V$ by $T(x)=Ax$ and $U(x)=A^*x$.
Show that $[U]_\beta={[T]_\beta}^*$ for any orthonormal basis $\beta$ for $V$.

Too tricky to solve!

Answer 1

In short, the question really is: prove that $[T^*]_\beta=[T]_\beta^\ast$ for every orthonormal basis $\beta$.

Let $\beta$ be an orthonormal basis of $V$. In general, for any linear operator $S:V\longrightarrow V$, the $(i,j)$ coefficient $a_{ij}$ of $[S]_{\beta}$ is $(Se_j,e_i)$. Indeed, $Se_j=\sum_{i=1}^na_{ij}e_i$ by definition of $[S]_\beta$. So it suffices to take the inner product of the latter equation with $e_i$ to obtain the claim.

Now $$(T^*e_j,e_i)=(e_j,Te_i)=\overline{(Te_i,e_j)}\qquad \forall 1\leq i,j\leq n$$ says precisely that the coefficients of $[T^\ast]_\beta$ and $[T]_\beta^\ast$ are the same.

Note: with the canonical basis $\beta_0$, you get that the matrix of $T$ is $A$, the matrix of $U$ is $A^*$. So $A^*=[U]_{\beta_0}=[T]_{\beta_0}^\ast=[T^\ast]_{\beta_0}$, whence $U=T^*$. So the question does amount to what I said in my first sentence.