Find a positive-definite matrix $B$ such that $\langle A,B\rangle_F=0$ Let $A$ denotes an $n\times n$ Hermitian matrix. Is there any positive-definite matrix, $B$, such that $\langle A,B\rangle_F=0$?

Such a positive definite $B$ exists if and only if $A$ is zero or indefinite.
If $B$ is positive definite and $A$ is semidefinite but nonzero, then $B^{1/2}AB^{1/2}$ is semidefinite but nonzero. Hence $\langle A,B\rangle_F=\operatorname{tr}(AB^\ast)=\operatorname{tr}(B^{1/2}AB^{1/2})\ne0$.
If $A$ is zero or indefinite, let $A=U\operatorname{diag}(x)U^\ast$ be a unitary diagonalisation. Then either $x$ is zero or $x$ has a pair of entries of different signs. Hence there exists a positive vector $y$ that is orthogonal to $x$. Let $B=U\operatorname{diag}(y)U^\ast$ and we are done.
