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Let $A$ - is some unital $C^*$ algebra, and $P$ is set of all strictly positive elements in $A$. We can define map $\sqrt{?} : P \to A$ which takes positive element and returns its (unique) strictly positive square root. How to evaluate its Frechet derivative?

Mykola Pochekai
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    I think you want $P $ to be the strictly positive elements. – Martin Argerami Jun 22 '16 at 01:36
  • Yes, I fix it. Thanks. – Mykola Pochekai Jun 22 '16 at 01:54
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    I suppose one could use term by term differentiation of Taylor series to solve this but I don't know if there is a more elegant explicit formula. – Jonas Meyer Mar 31 '17 at 14:01
  • An interesting reference I haven't checked thoroughly: http://eprints.ma.man.ac.uk/1218/01/covered/MIMS_ep2008_26.pdf "Computing the Fréchet Derivative of the Matrix Exponential..." by Al-Mohy and Higham. There is the observation in Theorem 2.1 which I had forgotten, $f\left(\begin{bmatrix}a&h\0&a \end{bmatrix}\right)=\begin{bmatrix}f(a) & Df(a,h)\ 0&f(a)\end{bmatrix}$. This sort of yields a formula: $Df(a,h)=\begin{bmatrix}1&0\end{bmatrix}\sqrt{\begin{bmatrix}a&h\0&a \end{bmatrix}}\begin{bmatrix}0\1\end{bmatrix}$. The authors use this to develop an iterative approximation for $Df$. – Jonas Meyer Apr 04 '17 at 02:17

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Let $P$ be the cone of strictly positive elements and let $V$ the the real linear combinations of these. $V$ is a real Banach space and $P$ is an open subset of $V$.

Consider the maps $s:P\to P, a\mapsto a^2$ and $r:P\to P,a\mapsto a^{1/2}$, you have that $s\circ r=\mathrm{id}\lvert_P=r\circ s$. It follows with the chain rule that

$$d(s\circ r)(a)=ds(r(a))\cdot dr(a) = \mathbb1\lvert_V=dr(s(a))\cdot ds(a)=d(r\circ s)(a)$$

So $dr(a)$ is the same as $ds(r(a))^{-1}$. Specifically $ds(r(a))=L(r(a))+R(r(a))$ where $L$ and $R$ are the left and right multiplication operations on $A$.

In the case of a commutative algebra the inverse becomes $\frac1{2\sqrt{a}}$, but I cannot find a nice expression in the general case, but I think this approach can be fruitful.

s.harp
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  • Yes, perhaps there isn't a nice formula that always holds and is as simple as those for integer powers. The observation in this answer was what I tried and had led me to experiment with weighted sums of maps of the form $h\mapsto a^{p}h a^{q}$ with no luck. Using a Taylor series for $\sqrt{}$ converging in a neighborhood of the spectrum of $a$ would allow a term by term derivative to be taken. A Cauchy integral formula I think might work, $h\mapsto \frac{1}{2\pi i}\int_C \sqrt{z}(zI-a)^{-1}h(zI-a)^{-1},dz$, where $C$ is in the right-half plane and encloses the spectrum of $a$. – Jonas Meyer Apr 04 '17 at 01:30
  • I don't know if there's a nice "explicit" formula that isn't a series, or an integral, or "the inverse of the derivative of squaring" (or some rewording of the latter). – Jonas Meyer Apr 04 '17 at 01:32
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    @JonasMeyer it appears that I received the bounty, even though this does not in my opinion qualify as an answer. I re-invested it into a second bounty of the same strength :) – s.harp Apr 06 '17 at 13:44