Derivative of Bures distance $\mbox{tr} \left( A + B - 2 \left( A^{1/2} B A^{1/2} \right)^{1/2} \right)$ with respect to matrix $A$ Let matrices $A$ and $B$ be symmetric and positive semidefinite (PSD). I'm trying to calculate the derivative of the Bures distance
$$d(A,B) := \mbox{tr} \left( A + B - 2 \left( A^{1/2} B A^{1/2} \right)^{1/2} \right)$$
with respect to $A$. I'm having issues with the last term $$2(A^{1/2}BA^{1/2})^{1/2}$$ involving matrix square roots. Does anyone know a nice way to compute this?
 A: As pointed out in the comments, for PSD matrices a drastic
simplification is possible:
$$\eqalign{
{\rm Tr}((A^{1/2}BA^{1/2})^{1/2}) &= {\rm Tr}((BA)^{1/2}) \\
}$$
In addition, there is a general result for the differential of the trace of any matrix function
$$\eqalign{
d\,{\rm Tr}\big(f(X)\big) &= f'(X^T):dX \\
}$$
where $f'$ is the ordinary derivative of the scalar function $f;\,$ both $f$ and $f'$ are evaluated using their respective matrix arguments.
Combining these yields a straightforward solution for the problematic term
$$\eqalign{
\phi &= {\rm Tr}\Big((BA)^{1/2}\Big) \\
d\phi
 &=  \tfrac 12\big((BA)^T\big)^{-1/2}:d(BA) \\
 &= \tfrac 12(AB)^{-1/2}:B\,dA \\
 &= \tfrac 12 B(AB)^{-1/2}:dA \\
\frac{\partial\phi}{\partial A}
 &= \tfrac 12 B(AB)^{-1/2} 
 \;=\; \tfrac 12 (BA)^{-1/2}B \\
}$$
Where the final equality is a theorem due to Higham
$$B\cdot f(AB) = f(BA)\cdot B$$
Therefore the gradient of the Bures Distance is
$$\eqalign{
\beta(A,B) &= {\rm Tr}\Big(A+B - 2(BA)^{1/2} \Big) \\
d\beta &= \Big(I - B(AB)^{-1/2}\Big):dA \\
\frac{\partial\beta}{\partial A}
 &= I - B(AB)^{-1/2} \;\;=\; I - (BA)^{-1/2}B \\
 &= I - A^{-1}(AB)^{1/2} \;=\; I - (BA)^{1/2}A^{-1} \\
}$$
All four gradient expressions are equivalent, and although it's not immediately obvious, the gradient is a symmetric matrix.
The gradient wrt $B$ can be derived in an analogous manner.
$$\eqalign{
\frac{\partial\beta}{\partial B}
 &= I - A(BA)^{-1/2} \;\;=\; I - (AB)^{-1/2}A \\
 &= I - B^{-1}(BA)^{1/2} \;=\; I - (AB)^{1/2}B^{-1} \\
}$$
