Is $tr(TA)=tr(TB)$ for any Hermitian operator $T$ equivalent to $A=B$? ($A,B$ are operators)

Question

I wonder whether this is true:

For any Hermitian operator $T$, tr($TA$)=tr($TB$), this is equivalent to $A=B$ ($A,B$ are operators).

I tried to prove it by contradiction, but failed to prove:

If $A\neq B$, then there exists a Hermitian $T$ such that tr($TA$)$\neq$tr($TB$).

Answer 1

It suffices to show that if for all Hermitian matrix $T$, $\text{tr}(TA)=0$, then $A=0$. We assume that the space is finite-dimensional so we can write $$A=(a_{jk})$$ with respect to an orthonormal basis.

Let's write $E_{jk}$ be the matrix with zero entries except the $(j,k)$-th entry is $1$. It is easy to see that $\text{tr}(E_{jk}A)=a_{kj}$.

Put $T=E_{jj}$. Then we have $\text{tr}(TA)=a_{jj}=0$.
Put $T=E_{jk}+E_{kj}$. Then $\text{tr}(TA)=a_{kj}+a_{jk}=0$.
Put $T=iE_{jk}-iE_{kj}$. Then $\text{tr}(TA)=ia_{kj}-ia_{jk}=0$.

Thus, all entries of $A$ is zero, contradiction.