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 $\langle A,B\rangle_F=0$?
 A: 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.
A: Assuming we are working with complex matrices here, the square of the Frobenius norm can also be written as $\text{tr}(A^\dagger B) = \text{tr}(A B)$, where $\dagger$ denotes the complex conjugate transpose.
Of course, if $A$ is positive semidefinite, then it's impossible for $\text{tr}(AB)$ to be zero unless $B$ or $A$ is the zero matrix, since $AB$ is also positive semidefinite.
But if $A$ is indefinite, then it's possible! If you unitarily diagonalize $A$ as $U D U^\dagger$ with $U$ unitary, and you choose $B$ to be simultaneously diagonalizable as $U E U^\dagger$, you can observe that
$$\langle A, B \rangle_F = \text{tr}(U D U^\dagger U E U^\dagger) = \text{tr}(U D E U^\dagger) = \text{tr}(DE)$$
So all you have to do is to find a positive diagonal matrix $E$ such that $D E$ has a zero trace. There are many ways to do this. One way is the following. Let $D$'s diagonal elements be $\lambda_1, \lambda_2, \cdots, \lambda_n$. WLOG let $\lambda_1 > 0$ and $\lambda_2 < 0$ (you can always rearrange the eigenvalues in the eigenvalue decomposition). Then let us choose $E$'s diagonal values (in order) to be
$$\begin{cases}
\displaystyle \left\{\frac{-\lambda_2}{\lambda_1} - \frac{1}{\lambda_1} \sum_{i = 3}^n \lambda_i, 1 , 1, \cdots, 1\right\},  & \text{if } \displaystyle \sum_{i = 3}^n \lambda_i < 0 \\
\displaystyle \left\{1, \frac{-\lambda_1}{\lambda_2} - \frac{1}{\lambda_2} \sum_{i = 3}^n \lambda_i, 1, 1, \cdots, 1\right\},  & \text{if } \displaystyle \sum_{i = 3}^n \lambda_i \geq 0
\end{cases}$$
I will leave it up to you to verify that all of the diagonal entries of $E$ constructed in this way are positive.
