Tensor Inverse (Optical Activity)

[Motocross9]

Homework Statement

Given $$\varepsilon^{'}=\varepsilon+i\epsilon_{0}[G]$$
$$(\varepsilon^{'}$$ is a hermitian, second rank tensor) show that
$$\varepsilon^{'-1}=\varepsilon^{-1}-i\epsilon_{0}\varepsilon^{-1}[G]\varepsilon^{-1}$$, note that $$[G]$$ is small. Also, $$\varepsilon$$ is a diagonal second rank tensor, and $$[G]$$ is a real antisymmetric matrix.

Relevant Equations

$$\varepsilon^{'}=\varepsilon+i\epsilon_{0}*[G]$$

Clearly, they used the binomial expansion on this; however, I cannot figure out why [G] is sandwiched by the epsilon inverses:
$$\varepsilon^{'-1}=1/(\varepsilon+i\epsilon_{0}[G])\approx(1-i\epsilon_{0}[G]\varepsilon^{-1})\varepsilon^{-1}$$

 

[jasonRF]

Hi Motocross9,

While most of us who are scientists or engineers use abusive (incorrect!) mathematical notation at times, in this case I think it is causing you to make mistakes in elementary matrix algebra. Using more proper notation, you first step is something like this
$$
\begin{eqnarray}
\left[ \epsilon + i \epsilon_0 G \right]^{-1} & = & \left[ \epsilon \left( I + i \epsilon_0 \epsilon^{-1} G \right) \right]^{-1}.
\end{eqnarray}
$$
What do you get if you continue from here?

jason

 

[Motocross9]

Hi Motocross9,

While most of us who are scientists or engineers use abusive (incorrect!) mathematical notation at times, in this case I think it is causing you to make mistakes in elementary matrix algebra. Using more proper notation, you first step is something like this
$$
\begin{eqnarray}
\left[ \epsilon + i \epsilon_0 G \right]^{-1} & = & \left[ \epsilon \left( I + i \epsilon_0 \epsilon^{-1} G \right) \right]^{-1}.
\end{eqnarray}
$$
What do you get if you continue from here?

jason

Thanks so much! I see my mistake now.