TITLE: Derivative (or differential) of symmetric square root of a matrix
QUESTION [13 upvotes]: Let A be a square, symmetric, positive-definite matrix.  Let S be its symmetric square root found by a singular value decomposition.  Let vech() be the half-vectorization operator.
Is there a convenient expression for the derivative (or differential) of vech(S) with respect to vech(A)?
I know the expression for an inverse, which is sort of like a matrix version of the power rule.  Would this approach work for a symmetric square root as well (i.e., (1/2)S^(-1/2))?

REPLY [5 votes]: To summarize Abhishek's post
$$\eqalign{
A &= S^2,\quad a={\rm vec}(A),\quad s={\rm vec}(S),\quad M=S^T\oplus S \cr
ds &= M^{-1}\,da \cr
}$$
However, to answer the original question, one must introduce Duplication and Elimination matrices.
$$\eqalign{
\alpha &= {\rm vech}(A),\quad \alpha=L_na,\quad a=D_n\alpha \cr
\sigma &= {\rm vech}(S),\quad \sigma=L_ns,\quad s=D_n\sigma \cr
L_n\,ds &= L_nM^{-1}(D_nL_n\,da) \cr
d\sigma &= L_nM^{-1}D_n\,\,d\alpha \cr
\frac{\partial\sigma}{\partial\alpha} &= L_nM^{-1}D_n \cr
}$$