Let $A$ and $B$ be two real square non-commuting matrices ($A,B \in \mathbb{R}^{N \times N}$ with $[A,B] = AB-BA \ne 0$). I also assume that $A$ is positive-definite. Consider the power function $$f : \mathbb{R}^{N \times N} \to \mathbb{R}^{N \times N}, \quad X \mapsto X^q. $$ When $q \in \mathbb{N}$ it is simply multiplying $X$ by itself $q$ times. I am interested in the case when $q \in \mathbb{Q}$ is fractional, e.g. $q = \frac{1}{2}$.

Although probably the easiest way to compute $(A+B)^q$ is by eigen-decomposition and applying $\lambda_i \mapsto \lambda_i^q$ on the eigenvalues, I want to use a different approach to calculate it by using Taylor expansion around $A$ where $B$ is sufficiently small (assume I know $A^{1/2}$). In this paper the following definition of Taylor expansion is given: $$ f(A+B) = \sum_{n=0}^{\infty} \frac{1}{n!} D_f^{[n]}(A,B) $$ where $$ D_f^{[n]}(A,B) = \left. \frac{d^n}{dt^n}\right|_{t=0} f(A+tB) $$ is the Frechet derivative.

This is the first time I encounter Frechet derivative. I try to read some about it (e.g. Wikipedia, or this paper, but they lack explicit examples). The last source offers an algorithm to compute them order by order, but I think it is beyond the the level of desired solution (which I want to implement in Python).

However I don't how to explicitly compute these derivatives for $f(X)=X^q$ when $q \in \mathbb{Q}-\mathbb{N}$, in particular for $q=1/2$. For the 1st derivative I tried to use the definition: $$ D_f^{[1]}(A,B) = \left( (A+tB)^q - A^q \right) + o(t) $$ but I don't know how to expand $(A+tB)^q$ around $A$ when $A$ and $B$ do not commute. When $A$ and $B$ commute then $$ D_f^{[n]}(A,B) = f^{(n)}(A) B^n $$ where $f^{(n)}(x) = \frac{d^n f(x)}{d x^n}$ is the scalar n-th derivative of $f$ (I assume $f$ is infinitely differentiable in $X = A$). I wonder if there is a closed form or formula to these derivatives when $f(X) = X^q$.

For the sake of completeness, when $q=1/2$ and $a,b \in \mathbb{R}^+$ are scalars then the Taylor expansion is $$ (a+b)^{1/2} = a^{1/2} + \sum_{n=1}^{\infty} \binom{1/2}{n} a^{\frac{1}{2}-n} b^n $$

Edit: the first Frechet derivative of $Y = X^{1/2}$ can be computed as follows: the Frechet derivative of $X = Y^2$ in $E$ is obtained via the definition $$ L_{y^2}(Y,E) = (Y+E)^2 - Y^2 = Y^2 + YE + EY + E^2 - Y^2 = YE + EY + o(\| E \|) $$ and since the Frechet derivative of the inverse of $X = Y^2$ is the inverse of $L_{y^2}(Y,E)$, that is: $$ L_{y^2}(Y,L_{x^{1/2}}(X,E)) = E $$ one need to solve the Sylvester equation $$ X^{1/2} L + L X^{1/2} = E $$ where $L = L_{x^{1/2}}(X,E)$ is the desired 1st Frechet derivative. There is a known algorithm to solve it and it is even implemented in Python's SciPy package/extension.

But how do I compute higher Frechet derivatives in this case ($X \mapsto X^q$ in particular for $q=1/2$)?

So to summarize my question(s):

  1. How to compute the Taylor expansion $(A+B)^q$ where $A > 0$ and $B$ is small but $[A,B] \ne 0$?
  2. If there is no closed form in the general case, is there one at least for $q=1/2$?
  3. How to compute the Frechet derivative of this Taylor expansion?

1 Answer 1


Assume that $A\in M_n$ is $>0$ real symmetric and $H\in M_n$ is a small real symmetric matrix. Let $q$ be a positive integer and $A^{1/q}=B$. We search an approximation of $(A+H)^{1/q}$ and we know $A,B,H$. Let $f:X\mapsto X^{1/q}$.

Then $f(A+H)=B+Df_A(H)+O(||H||^2)$, where $df_A(H)=K$ satisfies


If $q>2$, we are dealing with a generalized Sylvester equation in the unknown $K$. If $q=2$, this is the standard Sylvester equation -it admits a sole solution $K$ that is symmetric-. Now, we write about this last case

Several algo. can solve the equation $H=KB+BK$ with complexity $\approx 20n^3$.

$\textbf{Remark.}$ We can also diagonalize $A+H=PDP^T$; then

$f(A+H)=PD^{1/2}P^T$; the complexity is approximately the same as above.


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