Compute $\frac{d}{dx} \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ u\right\} $, where $M(x) = \left(B + xA \right) ; \ M^H(x) = M(x)$

Question

Problem:

Let $x \in \mathbb{R}$, $u \in \mathbb{C}^{n \times 1}$, $B \in \mathbb{C}^{n \times n}$, $A \in \mathbb{C}^{n \times n}$, and $M(x) = \left(B + xA \right) ; \ M^H(x) = M(x)$.

Obtain \begin{align} \frac{d}{dx} f(x) = \frac{d}{dx} \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ u\right\} \ . \end{align}

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Answer 1

Answer: \begin{align} \frac{d}{dx} \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ u\right\} &= -2 \ u^H M^{-1}(x) \ A \ M^{-1}(x) \ A \ M^{-1}(x) \ u \ . \end{align}

Solution:

Notation clarification: Complex conjugate of $u$ is denoted by $u^*$, and the conjugate transpose (or Hermitian) of $A$ is denoted by $A^H$.

I have utilized the following identities

• Trace and Frobenius product relation $${\rm tr}(A^H B) = A^* : B$$ or $${\rm tr}(AB) = A^H : B = (A^*)^T : B$$
• Cyclic property of Trace/Frobenius product \begin{align} A : B C &= AC^T : B \\ &= B^T A : C \\ &= {\text{etc.}} \cr \end{align}

So, we compute the differential first, and then the gradient.

Firstly, we compute the differential of $M^{-1}(x)$, i.e., $dM^{-1}(x)$, \begin{align} & I = M(x) M^{-1}(x) \\ & \Rightarrow 0 = dM(x) M^{-1}(x) + M(x)dM^{-1}(x) \\ & \Leftrightarrow dM^{-1}(x) = -M^{-1}(x) \ \underbrace{dM(x)}_{= A dx} M^{-1}(x) = - M^{-1}(x) \ A \ M^{-1}(x) \ dx \ . \end{align}

So, the differential of $f(x)$ reads \begin{align} df(x) &= d \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ u \right\} \\ &= d \left\{ u^* : M^{-1}(x) \ A \ M^{-1}(x) \ u \right\} \\ &= u^* : d \left\{M^{-1}(x) \ A \ M^{-1}(x) \ u \right\} \\ &= u^* : dM^{-1}(x) \ A \ M^{-1}(x) \ u + M^{-1}(x) \ A \ dM^{-1}(x) \ u \\ &= u^* : \left\{ \left[ - M^{-1}(x) \ A \ M^{-1}(x) \ dx \right] \ A \ M^{-1}(x) u \ + \ M^{-1}(x) \ A \ \left[ - M^{-1}(x) \ A \ M^{-1}(x) \ dx \right] u \right\} \\ &= -u : \left\{ \left[M^{-1}(x) \ A \ M^{-1}(x) \right] \ A \ M^{-1}(x) u \ + \ M^{-1}(x) \ A \ \left[ M^{-1}(x) \ A \ M^{-1}(x) \right] u \right\}^* \ dx \\ &= -\left\{ \left[M^{-1}(x) \ A \ M^{-1}(x) \right] \ A \ M^{-1}(x) u \ + \ M^{-1}(x) \ A \ \left[M^{-1}(x) \ A \ M^{-1}(x) \right] u \right\}^H u : \ dx \\ &= -\left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ A \ M^{-1}(x) u \ + u^H M^{-1}(x) \ A \ M^{-1}(x) \ A \ M^{-1}(x) u \right\} : dx \\ &= -2\left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ A \ M^{-1}(x) \ u \right\} : dx \end{align}

The gradient is \begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \ u\right\} \\ &= -2 \ u^H M^{-1}(x) \ A \ M^{-1}(x) \ A \ M^{-1}(x) \ u \ . \end{align}