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I have some issues in finding an asymptotic expansion for a square root matrix and I have already posted a question (Asymptotic expansion square root matrix). Somebody redirected me to a post where there is a possible answer to my question (How to calculate the square root of matrix $A+B$ perturbatively?). However I am not totally sure about such an answer and I would like to share my doubts.

Let formalize a little bit the definition of quantities we are dealing with in the latter post. The quantity we focused on may be thought to have an expansion in powers of $n^{-1/2}$ of the form $A+n^{-1/2}B+O(n^{-1})$ as $n\rightarrow\infty$, where $A=O(1)$ and $B=O(1)$ are symmetric positive definite matrices. Suppose that we want the same expansion for the square root matrix. Such a matrix is assumed to be symmetric (and obviously positive definite) as well and to have the expansion $C+n^{-1/2}D+O(n^{-1})$, with $C=O(1)$ and $D=O(1)$. By applying the identity $$(C+n^{-1/2}D)(C+n^{-1/2}D)=CC+n^{-1/2}CD+n^{-1/2}DC+O(n^{-1})=A+n^{-1/2}B+O(n^{-1})$$ and by matching the expansion term by term we are able to find an expansion for the square root matrix in terms of $A$ and $B$. In particular, $C$ and $D$ must fulfill the equations $A=CC$ and $B=CD+DC$. I hope that all of us would agree that $C=A^{1/2}$, so my point is the following: how it is possible that the solution of the second equation $B=A^{1/2}D+DA^{1/2}$ is $D=A^{-1/2}B/2$? I was told that this is a Lyapunov equation and for this kind of equation, I have not seen such a beauty and explicit solution in G. R. Duan and R. J. Patton (1998). Explicit and analytical solutions to Lyapunov algebraic matrix equations. UKACC International Conference on CONTROL 1998.

Any ideas?

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  • $\begingroup$ the solution to $B=\sqrt A D+D\sqrt A$ is given in your second link: $D=\int_0^\infty e^{-t\sqrt{A}}Be^{-t\sqrt{A}}dt$; this is a symmetric matrix. $\endgroup$
    – Carlo Beenakker
    Mar 5, 2015 at 20:24
  • $\begingroup$ Thank you Carlo again for the reply. What it is not clear to me is how from that integral we obtain $D=A^{-1/2}B/2$, or equivalently, the final expansion for the square root matrix $A^{1/2}+n^{-1/2}A^{-1/2}B/2$. $\endgroup$
    – user68601
    Mar 5, 2015 at 22:18
  • $\begingroup$ there is no closed-form expression for that integral, at least I do not know of any; given $A$ and $B$ you can evaluate it, but not in general. $\endgroup$
    – Carlo Beenakker
    Mar 5, 2015 at 22:57
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    $\begingroup$ Diagonalize, so that $\sqrt{A} = \mathrm{diag}(\sqrt{a_i})$. In the same basis, $B_{ij} = D_{ij} (\sqrt{a_i} + \sqrt{a_j})$, so that $D_{ij} = \frac{B_{ij}}{\sqrt{a_i} + \sqrt{a_j}}$, which is the unique solution and obviously symmetric. $\endgroup$
    – Igor Khavkine
    Mar 5, 2015 at 23:37
  • $\begingroup$ @Igor: thank you for the reply! Your solution involves the eigen values and vectors of $A^{1/2}$, Am I wrong? If not, regrettably your solution is not feasible for my purposes as it does not allow me to develop further calculations. $\endgroup$
    – user68601
    Mar 9, 2015 at 9:11

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