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i want to caculate inverse matrix, supposing initial matrix is in first order, and to expand inverse matrix in first order, but i dont know what $\frac{∂A^{-1}_{mn}}{∂a^{ij}}$ is。

i know that $\frac{dA^{-1}}{dt}=-A^{-1}\frac{dA}{dt}A^{-1}$

Is the following caculation correct?

Suppose $a^{ij}→a^{ij}+ɛ$，get $\frac{dA^{mn}}{dɛ}=1$ only if mn=ij

$$ \frac{∂A^{-1}_{mn}}{∂a^{ij}}=\frac{dA^{-1}_{mn}}{dɛ}=-A^{-1}_{mx}\frac{dA^{xy}}{dɛ}A^{-1}_{yn}=-A^{-1}_{mi}A^{-1}_{jn} $$

Exzample:

i'm studying general relativity, the easiest linear solution ( slightly curved spacetimes ) and Kerr metric in limit.

just write a metric like

$$ g_{μν}= \begin{pmatrix} 1+ɛ & & & j \\ & -1+ɛ & & \\ & & -1+ɛ & \\ j & & & -1+ɛ \end{pmatrix} ≈ \begin{pmatrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{pmatrix} $$

How to caculate $g^{μν}$ in first order? Is it $$ g^{μν}= \begin{pmatrix} 1-ɛ & & & j \\ & -1-ɛ & & \\ & & -1-ɛ & \\ j & & & -1-ɛ \end{pmatrix} $$

渡鸦100191
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  • The expression that you derived for $\frac{\partial A_{mn}^{-1}}{\partial A_{ij}}$ is correct. Not sure about the metric tensor. What is $j$ in those corner elements? – greg Aug 01 '21 at 23:40
  • i apologize for that i made a mistake of j, which is also a infinitesimal number like ɛ. ɛ = 2φ = -2GM/r in linear newtonian solution. j stands for something proportion to momenta of inertia, which should appear in spherical-coodinate Kerr metric, not this form in Cartesian coordinates ( in cartesian j should be in [0,1][0,2] not [0,3]). Here i just want to calculate something infinitesimal with no real physical meaning. so i mixed ɛ and j in the same linear metric. In spherical coordinates ɛ and j will also appear at the same time but the coefficients will not be 1/-1 – 渡鸦100191 Aug 03 '21 at 03:06

1 Answers1

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It is shown here: Prove if $\lim_{k \rightarrow \infty} A_k = A$, then $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$ that if $A \in GL(n, \mathbb{R})$ and $H \in M(n, \mathbb{R})$ with $\Vert H \rVert < \frac{1}{\lVert A^{-1} \rVert}$, then $$(A + H)^{-1} = A^{-1} - A^{-1}HA^{-1} + O(\lVert H \rVert^2).$$ Here $\lVert \cdot \rVert$ denotes the operator norm.

Mason
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