Find $\frac{d}{dx} \left( \lVert u - M^{-1}(x)u\rVert_2^2 \right)$, where $M(x) = \left(B + xA \right)$ and $M^H(x) = M(x)$

Question

Problem:

Let $x \in \mathbb{R}$, $u \in \mathbb{C}^{n \times 1}$, $B \in \mathbb{C}^{n \times n}; B^H = B$, and $A \in \mathbb{C}^{n \times n}; A^H = A$.

Find \begin{align} \frac{d}{dx} f(x) = \frac{d}{dx} \left\{ \left\| u - M^{-1}(x)u \right\|_2^2 \right\} \ , \end{align} where \begin{align} M(x) = \left(B + xA \right) \ \Rightarrow M^H(x) = M(x). \end{align}

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Answer 1

Incorporating feedback from greg, here is the answer (please no need to give credits).

Answer: \begin{align} \frac{d}{dx} \left\{ \left\| u - M^{-1}(x)u \right\|_2^2 \right\} &= 2 \ \Re \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \left[ u - M^{-1}(x) u \right] \right\} \ . \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}

Since \begin{align} f(x) &= \left\| u^* - M^{-1}(x) u \right\|_2^2 \\ &= \underbrace{\left[ u^* - M^{-*}(x) u^* \right]}_{= \left[ u^* - M^{-1}(x) u^* \right]} : \left[ u - M^{-1}(x) u \right] \\ &= \left[ u^* - M^{-1}(x) u^* \right] : \left[ u - M^{-1}(x) u \right] \end{align}

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

The gradient is \begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} \left\{ \left\| u - M^{-1}(x)u \right\|_2^2 \right\} \\ &= 2 \ \Re \left\{ u^H M^{-1}(x) \ A \ M^{-1}(x) \left[ u - M^{-1}(x) u \right] \right\} \ . \end{align}