Let $T\colon V\to V$ over an Inner Product Space $V$. Let $B$ be an orthogonal basis for $V$. Find the simplest connexion between $[T^*]_B$ and $([T]_B)^*$.
So I know that if $B$ was an orthonormal basis for $V$ we would have received: $$[T^*]_B=\langle T^*{b_i},b_j\rangle=\operatorname{adj}\langle T{b_i},b_j\rangle=\operatorname{adj}([T]_B)_{ji}$$ and therefore $\,[T^*]_B=([T]_B)^*$.
I am not sure how to find this relation using $B$ as an orthogonal basis. Do I just normalise the vectors?
Let $A = \{u_1,\dots,u_n\}$ denote the orthonormal basis obtained by normalizing the orthogonal basis $B = \{v_1,\dots,v_n\}$. Let $d_j = \|v_j\|$ for $j = 1,\dots,n$. Note that $v_j = d_j u_j$ for each $j = 1,\dots,n$.
The change of basis matrix from $B$ to $A$ is the diagonal matrix $$ P^B_A = \pmatrix{d_1 \\ & \ddots \\ && d_n}. $$ We note that \begin{align} [T^*]_B &= (P^B_A)^{-1}[T^*]_A P^B_A = (P^B_A)^{-1}[T]_A^* P^B_A \\ & = (P^B_A)^{-1}(P^B_A[T]_B (P^B_A)^{-1})^* P^B_A \\ & = (P^B_A)^{-1}(P^B_A)^{-*}[T]_B^* (P^B_A)^*P^B_A = (P^B_A)^{-2} [T]_B^* (P^B_A)^2. \end{align} In other words, if $M$ denotes the diagonal matrix whose diagonal entries are $M_{jj} = d_j^2 = \|v_j\|^2$, then $$ [T^*]_B = M^{-1} [T]_B^* M. $$