derivative of inverse matrix

Theorem 1.

Suppose $A$ is a square matrix depending on a real parameter $t$ taking values in an open set $I\mathrm{\subseteq }\mathrm{R}$. Further, suppose all component functions in $A$ are differentiable, and $A\mathit{}\mathrm{(}t\mathrm{)}$ is invertible for all $t$. Then, in $I$, we have

$$\frac{d{A}^{-1}}{dt}=-A^{-1}\frac{dA}{dt}{A}^{-1},$$

where $\frac{d}{d\mathit{}t}$ is the derivative.

Proof.

Suppose $a_{ij}(t)$ are the component functions for $A$, and $a^{jk}(t)$ are component functions for $A^{-1}(t)$. Then for each $t$ we have

$$\sum _{j=1}^n{a}_{ij}(t)a^{jk}(t)={\delta }_i^k$$

where $n$ is the order of $A$, and ${\delta }_i^k$ is the Kronecker delta symbol. Hence

$$\sum _{j=1}^n\frac{d{a}_{ij}}{dt}{a}^{jk}+a_{ij}\frac{d{a}^{jk}}{dt}=0,$$

that is,

$$\frac{dA}{dt}{A}^{-1}=-A\frac{d{A}^{-1}}{dt}$$

from which the claim follows. ∎

Title	derivative of inverse matrix
Canonical name	DerivativeOfInverseMatrix
Date of creation	2013-03-22 14:43:52
Last modified on	2013-03-22 14:43:52
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	7
Author	matte (1858)
Entry type	Theorem
Classification	msc 15-01