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Infinitesimal Matrix Change

Let B, A, and e be square matrices with e small, and define

{\hbox{\sf B}}\equiv{\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf e}}),
\end{displaymath} (1)

where I is the Identity Matrix. Then the inverse of ${\hbox{\sf B}}$ is approximately
{\hbox{\sf B}}^{-1}=({\hbox{\sf I}}-{\hbox{\sf e}}){\hbox{\sf A}}^{-1}.
\end{displaymath} (2)

This can be seen by multiplying
$\displaystyle {\hbox{\sf B}}{\hbox{\sf B}}^{-1}$ $\textstyle =$ $\displaystyle ({\hbox{\sf A}}+{\hbox{\sf A}}{\hbox{\sf e}})({\hbox{\sf A}}^{-1}-{\hbox{\sf e}}{\hbox{\sf A}}^{-1})$  
  $\textstyle =$ $\displaystyle {\hbox{\sf A}}{\hbox{\sf A}}^{-1}-{\hbox{\sf A}}{\hbox{\sf e}}{\h...
...ox{\sf e}}{\hbox{\sf A}}^{-1}-{\hbox{\sf A}}{\hbox{\sf e}}^2{\hbox{\sf A}}^{-1}$  
  $\textstyle =$ $\displaystyle {\hbox{\sf I}}-{\hbox{\sf A}}{\hbox{\sf e}}^2{\hbox{\sf A}}^{-1} \approx 1.$ (3)

Note that if we instead let ${\hbox{\sf B}}'\equiv{\hbox{\sf A}}+{\hbox{\sf e}}$, and look for an inverse of the form ${\hbox{\sf B}}'^{-1}={\hbox{\sf A}}^{-1}+{\hbox{\sf C}}$, we obtain

$\displaystyle {\hbox{\sf B}}'{\hbox{\sf B}}'^{-1}$ $\textstyle =$ $\displaystyle ({\hbox{\sf A}}+{\hbox{\sf e}})({\hbox{\sf A}}^{-1}+{\hbox{\sf C}...
...}}{\hbox{\sf C}}+{\hbox{\sf e}}{\hbox{\sf A}}^{-1}+{\hbox{\sf e}}{\hbox{\sf C}}$  
  $\textstyle =$ $\displaystyle {\hbox{\sf I}}+{\hbox{\sf A}}{\hbox{\sf C}}+{\hbox{\sf e}}({\hbox{\sf C}}+{\hbox{\sf A}}^{-1})\equiv {\hbox{\sf I}}.$ (4)

In order to eliminate the ${\hbox{\sf e}}$ term, we require ${\hbox{\sf C}}=-{\hbox{\sf A}}^{-1}$. However, then ${\hbox{\sf A}}{\hbox{\sf C}}=-{\hbox{\sf I}}$, so ${\hbox{\sf B}}{\hbox{\sf B}}^{-1}={\hbox{\sf0}}$ so there can be no inverse of this form.

The exact inverse of ${\hbox{\sf B}}'$ can be found as follows.

{\hbox{\sf B}}'={\hbox{\sf A}}+{\hbox{\sf e}}={\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}}),
\end{displaymath} (5)

{\hbox{\sf B}}'^{-1}=[{\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}})]^{-1}.
\end{displaymath} (6)

Using a general Matrix Inverse identity then gives
{\hbox{\sf B}}'^{-1}=({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}})^{-1}{\hbox{\sf A}}^{-1}.
\end{displaymath} (7)

© 1996-9 Eric W. Weisstein