5
$\begingroup$

$A=diag\{\lambda_1,...,\lambda_n\}$ 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?

$\endgroup$
5
  • 1
    $\begingroup$ $\sqrt{A+B}=\sqrt{A}(1+A^{-1}B/2)$ $\endgroup$ – Carlo Beenakker Jan 14 '15 at 12:27
  • 2
    $\begingroup$ @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 Jan 14 '15 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 Jan 14 '15 at 12:42
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/196448/2451 $\endgroup$ – Qmechanic Aug 7 '15 at 19:25
  • 1
    $\begingroup$ What does it mean to "calculate something perturbatively"? $\endgroup$ – Suvrit Oct 25 '15 at 19:21
5
$\begingroup$

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 $A$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ The assumption of continuity of the derivative is unnecessary. $\endgroup$ – J Tyson Oct 25 '15 at 20:44
  • 1
    $\begingroup$ The link is broken. $\endgroup$ – Alexander Mathiasen Aug 10 '20 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.