# Matrix representation of adjoint operators

I am studying linear algebra. In the textbook I use, the book says that(Friedberg, 5th ed, p.356 theorem 10) If $$V$$ is an f.d.i.p.s and $$B$$ is an O.N.B for $$V$$, $$[T^{*}]_{B}=[T]_{B}^{*}$$ holds for every linear operator on $$V$$. However, the book doesn’t give any explanations about the inverse of this proposition. (If $$[T^{*}]_{B}=[T]_{B}^{*}$$ holds for all linear operator $$T$$, is B an O.N.B?) I found out that if $$B$$ is an Orthogonal(Not necessarily orthonormal) basis for $$V$$, then [$$[T^{*}]_{B}=[T]_{B}^{*}$$ iff all vectors in B has the same norm]. It can be done by using the definition and showing that $$||v_{i}||=||v_{j}||$$ for all elements in $$B$$.

My question is, what kind of bases satisfy $$[T^{*}]_{B}=[T]_{B}^{*}$$ for all linear operator $$V$$? The condition should comprise all orthonormal bases, all otrhogonal and same-norm bases. Can other bases satisfy it?

My question is, what kind of bases satisfy $$[T^{*}]_{B}=[T]_{B}^{*}$$ for all linear operator $$V$$? The condition should comprise all orthonormal bases, all otrhogonal and same-norm bases. Can other bases satisfy it?
Fix a basis $$B=\{v_1,\ldots,v_n\}$$. Let $$T$$ be the linear operator such that $$Tv_1=v_2$$, $$Tv_2=v_1$$, $$Tv_k=v_k$$ for $$k=3,\ldots,n$$. If $$[T^*]_B=[T]_B^*$$, since the latter is real and symmetric we get $$[T^*]_B=[T]_B^*=[T]_B.$$ It follows that $$T^*=T$$. Now $$\|v_2\|^2=\langle v_2,v_2\rangle=\langle Tv_1,v_2\rangle=\langle v_1,Tv_2\rangle=\|v_1\|^2.$$
Using appropriate variations of $$T$$ we get that $$\|v_j\|=\|v_k\|$$ for all $$k,j$$.
Now consider instead $$T$$ such that $$Tv_1=v_2$$ and $$Tv_k=0$$ for $$k=2,\ldots,n$$. From $$[T^*]_B=[T]_B^*$$ we get that $$T^*v_2=v_1$$ and $$T^*v_1=0$$. Then $$\langle v_2,v_1\rangle=\langle Tv_1,v_1\rangle=\langle v_1,T^*v_1\rangle=0.$$ Using appropriate variations of $$T$$ we see that $$B$$ is orthogonal.