Let A be an $n\times n$ matrix (real valued), and set $S := \sqrt{A^t A}$. What is the matrix derivative of $S$ w.r.t. $A$? I.e., what is $\frac{\partial S}{\partial A}$?
(In case of invertibility issues, I might for the time being assume that $A^t A$ is positive definite, although for the applications I have in mind I might have only positive semi-definiteness).
Thanks for help!
On
If you are familiar with the vec operation for matrices, then you could proceed as follows. $$\eqalign{ &S\,S &= A^TA \cr &S\,dS\,(I)+(I)\,dS\,S &= A^T\,dA\,(I)+(I)\,dA^T\,A \cr &(I^T\otimes S+S^T\otimes I)\,{\rm vec}(dS) &= (I^T\otimes A^T)\,{\rm vec}(dA)+(A^T\otimes I)\,{\rm vec}(dA^T) \cr &\Big(I\otimes S+S\otimes I\Big)\,{\rm vec}(dS) &= \Big((I\otimes A^T)+(A^T\otimes I)K\Big)\,{\rm vec}(dA) \cr &\frac{\partial{\,\rm vec}(S)}{\partial{\,\rm vec}(A)} &= \Big(I\otimes S+S\otimes I\Big)^+ \Big((I\otimes A^T)+(A^T\otimes I)K\Big) \cr\cr }$$ where $M^+$ denotes the pseudoinverse of $M$, $I$ is the identity matrix, and $K$ is the commutation matrix associated with the Kronecker product. The solution also takes advantage of the fact that $I$ and $S$ are symmetric.
Recall that you can perform a polar decomposition on $A$ such that:
$$A = QS$$
where $Q$ is an orthogonal matrix and $S$ is defined uniquely as precisely what you describe above when it exists. Hence, what you really want to calculate is:
$$\frac{\partial S}{\partial A} = \frac{\partial [Q^{-1}A]}{\partial A} = Q^{-1}\otimes\mathbf{I}$$
Note the result is a fourth-order linear transformation, since we are the describing the rate of change of a second-order object $S$ with respect to another second-order linear transformation $A$.