Derivative of Product of Matrix with Moore-Penrose Inverse

Question

Assume that $A(x) \in M_{m,n}$ is given and depends on some $x \in \mathbb{R}$. Let $A^\dagger(x) = [A(x)^\star A(x)]^{-1} A^\star(x)$ be the Moore-Penrose pseudoinverse of $A(x)$.

Define $G(x) = A(x) A^\dagger(x)$ and derive the derivative $\frac{\partial G}{\partial x}$.

My steps so far (trying to exploit that $\frac{\partial G}{\partial x} = \frac{\partial \text{vec}\ G(x)}{\partial x}$: $$ G(x) = A(x) [A(x)^\star A(x)]^{-1} A^\star(x) \\ \iff \text{vec}(G(x)) = \text{vec}(A(x) [A(x)^\star A(x)]^{-1} A^\star(x)) \\ = (A(x) \otimes A(x)) \text{vec}([A(x)^\star A(x)]^{-1}) $$ where $C \otimes D$ is the Kronecker product of $C$ and $D$.

Answer 1

Let $B=\frac{dA}{dx}$. By the product rule and the rule that $dY^{-1}=-Y^{-1}(dY)Y^{-1}$, we obtain \begin{aligned} \frac{dG}{dx}=\frac{dAA^\dagger}{dx} &=BA^\dagger+A\left[-(A^\ast A)^{-1}(B^\ast A+A^\ast B)(A^\ast A)^{-1}A^\ast+(A^\ast A)^{-1}B^\ast\right]\\ &=BA^\dagger-(A^\dagger)^\ast B^\ast AA^\dagger - (A^\dagger)^\ast A^\ast BA^\dagger+ (A^\dagger)^\ast B^\ast\\ &=BA^\dagger-(A^\dagger)^\ast B^\ast G - G^\ast BA^\dagger+ (A^\dagger)^\ast B^\ast\\ &=(I-G)^\ast BA^\dagger+(A^\dagger)^\ast B^\ast(I-G)\\ &=(I-G)BA^\dagger+(A^\dagger)^\ast B^\ast(I-G)^\ast\\ &=(I-G)BA^\dagger+\left[(I-G)BA^\dagger\right]^\ast. \end{aligned}

Sanity checks:

• As $G$ is Hermitian, $\frac{dG}{dx}$ should also be Hermitian.
• As $GA=AA^\ast A=A$, we have $\frac{dG}{dx}A+G\frac{dA}{dx}=\frac{dA}{dx}$. Hence $\frac{dG}{dx}A$ should be equal to $(I-G)B$.

Answer 2

If any solutions of the linear system $BA=C$ exist, they are given by $$B = CA^+ + R(I-G)$$ where $G=AA^+,\,$ $I$ is the identity matrix, and $R$ is an arbitrary matrix.

For ease of typing, use $\,dA=\frac{dA}{dx}$
Differentiating the first Penrose condition yields
$$\eqalign{ A &= GA \\ dA &= d(GA) = dG\,A + G\,dA \\ dG\,A &= (I-G)\,dA \\ dG &= (I-G)\,dA\,A^+ + R(I-G) \\ }$$ Since $(dG,I,G)$ are hermitian, we can identify $R$ to complete the solution. $$\eqalign{ dG^* &= dG \quad\implies\quad R^* = dA\,A^+ \\ dG &= (I-G)\,dA\,A^+ + (dA\,A^+)^*(I-G) \\ }$$