Given two positive self-adjoint operators $A,B$ on a Hilbert space. Let $p\geq 1$.

I would like to calculate $$\frac{d}{dt}|_{t=0} (A+tB)^p,$$

where the power is defined through the spectral theorem. The difficulty is that $A$ and $B$ do not commute in general. Is there a nice formula for that? I thought about using some integral representation of $x^p$, which hopefully would make the differentiation easier, but I was not successful.

Update: Thanks for your answers. In my case, I actually only need it for $p\in [1,2)$. Using $y^p = \frac{\sqrt 3}{2\pi}\int_{0}^\infty dx \,\frac{y^2 x^{p-2}}{x+y}$, I thought about writing $C:=A+tB$ as $$C^p = \frac{\sqrt 3}{2\pi}\int_0^\infty C^2 x^{p-2} (x+C)^{-1} dx.$$ That might also work. I am still happy about other suggestions.


In general, one standard approach would be to consider the map $A \mapsto A^p$ for any $p$. Then, actually write the map as $A \mapsto \exp(p \log A)$. Then, the derivative that you are after is the Fréchet derivative in direction $B$. This does not look entirely clean, but at this point it is just the chain rule (e.g., see the derivative of the matrix exponential here, and of the matrix logarithm (see Slide 38); also in the same slides, you'll see numerical procedures for computing these derivatives numerically efficiently).

Alternatively, check out the Daleckii-Krein theorem, which will show you how to obtain "nice" closed forms for derivatives of matrix functions similar to those in the above question.


I give you the answer in the finite dimensional case, and let you adapt it to the Hilbert space case. Let $f$ be an analytic function; in your case $f(t)=t^p$.

First of all, choose an orthonormal basis in which $A=:D$ is diagonal. Then form the matrix $f^{[1]}(D)\in{\bf M}_n({\mathbb C})$ by $$f^{[1]}(D)_{jk}=\left\{\begin{array}{lr} f'(d_j), & k=j, \\ \\ \frac{f(d_j)-f(d_k)}{d_j-d_k}, & k\ne j,\end{array}\right.$$ where we identify $$\frac{f(b)-f(a)}{b-a}:=f'(a),$$ if $b=a$. Then the differential of $H\mapsto f(H)$, taken at $D$, is $${\rm D}f(D)B=f^{[1]}(D)\circ B, $$ where $A\circ B$ denotes the Hadamard product.

Edit. As noticed by Suvrit, this is Daleckii-Krein Formula.

  • $\begingroup$ This is essentially the content of the Daleckii-Krein theorem :-) $\endgroup$
    – Suvrit
    Oct 22 '15 at 13:31
  • $\begingroup$ @Suvrit. Thanks. I didn't remember the name of this formula. $\endgroup$ Oct 22 '15 at 14:32

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