Matrix representation of adjoint operators

Question

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?

Answer 1

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?

No, only the bases you described do.

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.