For a given Hermitian matrix $A$, what is the semidefinite positive matrix $B$ such that $$\mathbf{y}^{H}B\mathbf{y} = \left | \mathbf{y}^{H}A\mathbf{y} \right |, $$ for all $\mathbf{y} \in \mathbb{C}^{n}$? The operation $|\cdot|$ is the absolute value.
My thoughts:
If $A$ is semidefinite positive, then $\left | \mathbf{y}^{H}A\mathbf{y} \right | = \mathbf{y}^{H}A\mathbf{y}$ and $B = A$. But, if the matrix A is indefinite, I don't know if it is possible to construct a matrix $B$ from $A$. I have just tried to change the negative eigenvalues of $A$ for $-\lambda$ and construct the matrix $B$ as follows $$A = UDU^{-1} \Rightarrow B = U\left |D\right| U^{-1},$$ where $\left |D\right|$ is the elementwise absolute value of $D$, but this certainly does not work.
Any ideas or hints to solve this problem? Thanks in advance!
I don't think this is possible. The problem is that while the quadratic form $y^*Ay$ is perfectly smooth for any Hermitian $A$, $|y^*Ay|$ does not need to be if $A$ is indefinite. Consider $$ f(t):=y(t)^TAy(t):=\pmatrix{t\\1-t}^T\pmatrix{1&0\\0&-1}\pmatrix{t\\1-t}=2t-1. $$ Clearly, $|f|$ is not differentiable at $t=1/2$.