This comes from a physical context (see density matrices), but I don't believe that is crucial to the problem.

Suppose I have the following $2$ x $2$ Hermitian matrix $$A=\begin{pmatrix} a & b \\ b^* & c \\ \end{pmatrix}$$ with the conditions that $Tr[A]=1$. Assuming I know the values of $a$ and $c$, what values of $b$ ensure that the resulting matrix is positive semidefinite?

I have seen from Ensuring that a symmetric matrix with nonnegative elements is positive semidefinite that it sufficient to have a diagonally dominant matrix with nonnegative terms on the diagonal, but I'm still wondering if I can find a necessary and sufficient condition to ensure I have a proper density matrix.


So $$A=\begin{pmatrix} a & b \\ b^* & 1-a \\ \end{pmatrix}.$$ Its eigenvalues are the soultions of the characteristic equation $0 = \det(A-\lambda E) = (a-\lambda)(1-a-\lambda)-|b|^2 = \lambda^2 - \lambda + a(1-a)-|b|^2 = \lambda^2 - \lambda + \det A. $

Both roots are non-negative iff $\det A\geqslant 0$, i.e., $\sqrt{a(1-a)}\geqslant|b|$.

Finally, the matrix is positive semidefinite when all eigenvalues are non-negative.


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