$A=\text{diag}\{\lambda_1,...,\lambda_n\}$, where $\lambda_i$ can be any number and not necessarily a small number, $\lambda_i>0$, $B$ is a positive definite symmetric matrix, and $\text{max}\{B_{ij} \}\ll \text{min}\{\lambda_i\}$.

Note that the perturbative calculation of square root of $I+B$ is very easy, where $B$ is a small matrix.
$$\sqrt{(I+B)}= I+\frac{1}{2}B-\frac{1}{8}B^2\cdots$$

In general how to calculate the square root of $A+B$ perturbatively?

This question is nontrivial because $\sqrt{A}\sqrt{B}\not=\sqrt{B}\sqrt{A}$, $\sqrt{AB}\not=\sqrt{A}\sqrt{B}$ and $\sqrt{AB}\not=\sqrt{B}\sqrt{A}$. If we write $C=A+B-I$ and $A+B=I+C$, $C$ is not a small quantity, therefore the perturbative expansion fails.

  • $\begingroup$ $A + B = A\left(I + A^{-1}B\right)$ $\endgroup$ Commented Jul 28, 2015 at 16:15
  • 2
    $\begingroup$ @CountIblis $\sqrt{A}\sqrt{B}\not=\sqrt{B}\sqrt{A}$ and $\sqrt{AB}\not=\sqrt{A}\sqrt{B}$ $\endgroup$
    – 346699
    Commented Jul 28, 2015 at 16:20
  • 1
    $\begingroup$ Write $A+B = I + C$, where $C = A + B - I$ and check that $C$ satisfy your criterion of being small. $\endgroup$
    – Phoenix87
    Commented Jul 28, 2015 at 16:23
  • $\begingroup$ @Phoenix87 In this case $C$ is not a small quantity, the perturbative expansion fails. $\endgroup$
    – 346699
    Commented Jul 28, 2015 at 16:26
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    $\begingroup$ $A+B$ is positive definite, so you can diagonalise this matrix and the entries will be perturbations of the eigenvalues of $A$. Hence if these are not all close to 1 (that is, if $A$ is not close to the identity matrix) I don't really see how to apply the perturbative method the OP refers to. $\endgroup$
    – Phoenix87
    Commented Jul 28, 2015 at 16:32

2 Answers 2



  1. The square root function has a Taylor expansion around $a>0$ $$ \sqrt{a+b}~=~\sum_{n=0}^{\infty} \begin{pmatrix}\frac{1}{2} \cr n\end{pmatrix}a^{\frac{1}{2}-n}b^n, \qquad |b| ~<~a. \tag{1} $$

  2. One may show that a possible non-commutative generalization reads $$\begin{align} \sqrt{A+B}~=~&\sqrt{A}+\sum_{n=1}^{\infty} \begin{pmatrix}\frac{1}{2} \cr n\end{pmatrix}\left( \left(L_B +{\rm ad} A \right)^{n-1}B \right) A^{\frac{1}{2}-n}, \cr A~>~&0, \end{align} \tag{2}$$ and $B$ sufficiently small. Here $$ L_B(A):=BA \quad\text{and}\quad ({\rm ad} A)B~:=~[A,B] \tag{3}$$ are left composition and adjoint action, respectively.

  • 1
    $\begingroup$ How to prove this claim? Or where can I find the proof? $\endgroup$
    – 346699
    Commented Aug 7, 2015 at 2:51

An elementary way to proceed is as follows. Let's put an explicit factor of $\epsilon$ in $B$. The problem is then to solve the following equation for the matrix $X$:

$$A + \epsilon B = X^{2}$$

We want to do this perturbatively, so we assume that $X$ can be represented as:

$$X = \sum_{n=0}^{\infty}\epsilon^{n}X^{(n)}$$

We can then write:

$$ \begin{split} X^2 &= \sum_{n=0}^{\infty}\epsilon^{n}\sum_{k=0}^{n}X^{(k)}X^{(n-k)}= \left(X^{(0)}\right)^2 + \epsilon\left(X^{(0)}X^{(1)}+X^{(1)}X^{(0)}\right) \\&+\sum_{n=2}^{\infty}\epsilon^{n}\left(X^{(0)}X^{(n)}+X^{(n)}X^{(0)} + \sum_{k=1}^{n-1}X^{(k)}X^{(n-k)}\right) \end{split} $$

We then see that:

$$X^{0} = A^{\frac{1}{2}}$$

In components we can write this as:

$$X^{(0)}_{i,j} = \sqrt{\lambda_{i}}\delta_{i,j}$$

where the summation convention is modified so that contraction only happens when that would also happen absent any factors of the $\lambda$'s in an equation.

We can then simplify the expression $Y^{(n)} \equiv X^{(0)}X^{(n)} + X^{(n)}X^{(0)}$:

$$Y^{(n)}_{i,j} = \left(\sqrt{\lambda_{i}}+\sqrt{\lambda_{j}}\right)X^{(n)}_{i,j}$$

We thus have:

$$X^{(1)}_{i,j} = \frac{B_{i,j}}{\sqrt{\lambda_{i}}+\sqrt{\lambda_{j}}}$$

And the higher order terms are then obtained recursively from:

$$X^{(n+1)}_{i,j} = -\frac{\sum_{r=1}^{n}X^{(r)}_{i,k}X^{(n+1-r)}_{k,j}}{\sqrt{\lambda_{i}}+\sqrt{\lambda_{j}}}$$


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