Derivative of a matrix given with a root

Question

Let us defined the operator $\boldsymbol{N}(x)=\sqrt{\boldsymbol{M}^{\dagger-1}(x)\,\boldsymbol{M}^{-1}(x)}$.

assuming that we know $\frac{d\boldsymbol{M}(x)}{dx}$, how do I compute $\frac{d\boldsymbol{N}(x)}{dx}$?

Answer 1

As greg hint indicates, we can proceed with a kind of implicit differentiation. Let $M'$ denote the derivative $M' = dM/dx$. Note that $$ M^\dagger N^2M = I. $$ Differentiating on both sides yields $$ [M']^\dagger N^2 M + M^\dagger N^2 M' + M^\dagger(NN' + N'N)M = 0 $$ Rearranging the equation gives us $$ M^\dagger(NN' + N'N)M = -[M']^\dagger N^2 M - M^\dagger N^2 M' \implies\\ NN' + N'N = -[M^{-1}]^\dagger[M']^\dagger N^2 - N^2 M'M^{-1}, $$ which is a Sylvester equation that can be solved for $N'$. The fact that $N$ is positive definite ensures that the solution to this equation is unique.

In fact, this is not just a Sylvester equation, it's more specifically a (continuous time) Lyapunov equation. As such, its solution can be given explicitly as the result of a certain integral.

In particular, if we denote $$ Q = -[M^{-1}]^\dagger[M']^\dagger N^2 - N^2 M'M^{-1}, $$ then the equation for $N'$ above may be written in the form $$ (-N)N' + N'(-N) + Q = 0, $$ so that $$ N'(x) = \int_{0}^\infty e^{-yN(x)}Qe^{-yN(x)}\,dy. $$