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$A=diag\{\lambda_1,...,\lambda_n\}$ and $\lambda_i>0$, $B$ is a positive definite symmetric matrix and $max\{B_{ij} \}\ll min\{\lambda_i\}$

Note that the perturbative calculation of square root of $I+B$ is very easy, where $B$ is a small matrix.

How to calculate the square root of $A+B$ perturbatively?

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    $\begingroup$ $\sqrt{A+B}=\sqrt{A}(1+A^{-1}B/2)$ $\endgroup$
    – Carlo Beenakker
    Commented Jan 14, 2015 at 12:27
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    $\begingroup$ @CarloBeenakker This answer is not right. Because $\sqrt{A}(1+A^{-1}B/2)\sqrt{A}(1+A^{-1}B/2)- (A+B)$ is not a second order quantity. $\endgroup$
    – 346699
    Commented Jan 14, 2015 at 12:31
  • $\begingroup$ What kind of "second order" do you want? $O(\|B\|^2)$ independently of $A$ might be difficult to achieve, since for instance we could have $A=0$. $\endgroup$
    – Federico Poloni
    Commented Jan 14, 2015 at 12:42
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/196448/2451 $\endgroup$
    – Qmechanic
    Commented Aug 7, 2015 at 19:25
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    $\begingroup$ What does it mean to "calculate something perturbatively"? $\endgroup$
    – Suvrit
    Commented Oct 25, 2015 at 19:21

2 Answers 2

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EDIT. Finally user34669 is right. We assume that $A$ is fixed and $B$ tends to $0$. The following $3$ lines are not correct because, in general, $\sqrt{XY}\not=\sqrt{X}\sqrt{Y}$.

"$\Delta=\sqrt{A+B}-\sqrt{A}\approx (A^{-1/2}B)/2$. We may also write $\Delta\approx (BA^{-1/2})/2$ or in a symmetric form $\Delta\approx (A^{-1/2}B+BA^{-1/2})/4$ and the formula is valid for $A$ symmetric $>0$ and $B$ a small symmetric matrix."

In fact, the previous approximations give an error in $O(||[A^{1/2},B]||)$, that is not interesting except if $A,B$ commute (we would like an error in $O(||B||^2)$ or at least in $o(||B||)$).

On the other hand, the other part is correct.

The function $S:X\rightarrow \sqrt{X}$ is defined and derivable on the set of SPD matrices. Let $K=DS_A(H)$ be the derivative of $S$ in $A$, where $H$ is a variable SYMMETRIC matrix. Here $SS=I$ implies $K\sqrt{A}+\sqrt{A}K=H$, a Sylvester equation in the unknown $K$.

i) A closed form. It is known that $K=\int_0^{\infty}e^{-t\sqrt{A}}He^{-t\sqrt{A}}dt$. That implies that if $B$ is a small symmetric matrix, then $\Delta\approx \int_0^{\infty}e^{-t\sqrt{A}}Be^{-t\sqrt{A}}dt$, a symmetric matrix.

ii) Numerically, it is easier to solve directly the Sylvester equation diagonalizing $A$.

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See Theorem 3.25 on page 122 (as given by the page numbers) of http://www.renyi.hu/~petz/pdf/matrixPD.pdf

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  • $\begingroup$ The assumption of continuity of the derivative is unnecessary. $\endgroup$
    – J Tyson
    Commented Oct 25, 2015 at 20:44
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    $\begingroup$ The link is broken. $\endgroup$
    – Alexander Mathiasen
    Commented Aug 10, 2020 at 14:11

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