Dear Matrix Calculus Experts,

Let $n\geq p$ and $x\in\mathbb{R}\rightarrow A(x)\in G_{n,p}(\mathbb{C)}$ be a $C^1$ function where $G_{n,p}$ is the subset of $M_{n,p}$ constituted by the matrices with rank $p$.

Can we differentiate the following function $f:x\rightarrow A(x)(A(x)^*A(x))^{-1} A(x)^*$ ?

So, we need to obtain (Jacobian matrix?), i.e., $$ \frac{\partial}{\partial x} \left\{ A(x) \left(A(x)^*A(x)\right)^{-1} A(x)^* \right\}$$ in terms of $ \frac{\partial}{\partial x} A(x)$.

Thank you very much in advance.

  • $\begingroup$ You could make the effort to say what $A(x)$ is. $\endgroup$ – loup blanc May 12 '18 at 21:02
  • $\begingroup$ @loupblanc it is not given $\endgroup$ – user550103 May 13 '18 at 5:57

Let $M$ denote the Moore-Penrose inverse of $A$, and for ease of typing let $$dA = \frac{dA}{dx}$$ The function you are interested in can be written $$f = AM$$ The derivative of $M$ (in a direction of constant rank) is a well known result $$dM = -M(dA)M + MM^*(dA^*)(I-AM) + (I-MA)(dA^*)M^*M$$ Therefore the derivative of your function is $$\eqalign{ df &= d(AM) \cr &= (dA)M + A(dM) \cr &= (dA)M -AM(dA)M + M^*(dA^*)(I-AM) + 0 \cr &= (I-AM)(dA)M + \Big((I-AM)(dA)M\Big)^* \cr }$$

  • $\begingroup$ Thank you very much. When you say "well known" result, then can you give me one reference please? $\endgroup$ – user550103 May 13 '18 at 16:52

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