Positive definite matrix for complex vectorial space

Question

I have problems when building a positive definite matrix on a complex vectorial space. I came across this problem when testing an optimization algorithm, despite its theory only considers the real case.

This is my situation:

Let $\tau, \sigma \in \mathbb{R}$ and $x, y, k_1, k_2 \in \mathbb{C}$. Also $\tau, \sigma > 0$

I have this matrix: $$M = \begin{bmatrix} \frac{1}{\tau} & -k_2 \\ -k_1 & \frac{1}{\sigma} \end{bmatrix}$$

and I have this vector $$v = \begin{bmatrix} x \\ y \end{bmatrix}$$ In this context, I want to find values for $\tau$ and $\sigma$ that make $M$ positive definite with respect to $v$ ($v^H$ is its Hermitian transform). When I make the calculations I get: $$v^H Mv = \frac{x^Hx}{\tau} + \frac{y^Hy}{\sigma} - x^Hk_1y - y^Hk_2x > 0$$ $$\frac{x^Hx}{\tau} + \frac{y^Hy}{\sigma} > x^Hk_1y + y^Hk_2x$$ Here I have to compare a real value (to the left of the equation) and a complex value (to the right). I read somewhere that this kind of comparison is not defined in the complex case, even matlab considers only the real part.

I'm stuck here. Is it a correct approach to take the norm of both sides? $$\frac{\|x^Hx\|}{\tau} + \frac{\|y^Hy\|}{\sigma} > \|x^Hk_1y + y^Hk_2x\|$$

Answer 1

I think that taking the norms is a valid approach. The inequality stays valid, but I'm not sure that you will obtain the tightest possible bounds on $\tau$ and $\sigma$.

Edit: Since the LHS will be a scalar, I think you may need to place the norm outside the entire LHS expression.

Answer 2

Your conditions are $u=\overline{x}yk_1+x\overline{y}k_2\in \mathbb{R}$ and $w=||x||^2/\tau+||y||^2/\sigma>u$.

Let $v=x\overline{y}(k_2-\overline{k_1})$; one can see that $v\in\mathbb{R}$ IFF $u\in\mathbb{R}$.

If $k_2-\overline{k_1}=re^{i\theta}$, then necessarily $(*)$ $x\overline{y}$ is in the form $se^{-i\theta}$. If it's not the case, then no solutions.

If it's the case, then, to satisfy the second inequality, it suffices to choose $\tau,\sigma$ positive and small enough.