Determinant expansion I've seen in a few textbooks now a useful looking expansion procedure for determinants, but I don't understand the details of it. Here is precisely the example I'm thinking of (Ex. 7.6). I don't understand the notation
\begin{equation}\det(M+\Lambda)=\det(M)\left[\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{i_{1},i_{2}...i_{n}} \lambda_{i_{1}}\lambda_{i_{2}}...\lambda_{i_{n}}\det(\Delta_{i_{l}i_{k}})\right]\end{equation}
$\Delta=M^{-1}$ Does this mean $\det(\Delta_{i_{l}i_{k}})$ is some sort of co-factor determinant? Also, this was done in the context of Grassmann variables, so would that indicate that $M$ must be antisymmetric? Or does this hold for any $M$ (invertible)?
If someone could walk me through the notation of this I would be very grateful.
EDIT: I think my main source of confusion is what the author means by $\det(\Delta_{i_{l}i_{k}})$.
 A: The symmetric part of $M$ would just not contribute to the integral, but wouldn't affect the identity. I think you want to calculate $\det (1 + M^{-1} \Lambda)$ by doing a Grassmann integral over the variables $\theta_i$, i.e.,
$\int \prod_i d\theta_i \prod_j d\bar\theta_j   \exp [ \bar \theta^T (1 + M^{-1} \Lambda ) \theta]$,
or just
$\int \prod_i d\theta_i  \prod_j d\bar\theta_j  \exp [ \bar \theta^T (M + \Lambda ) \theta]$.
You can expand the exponential as a Taylor series, each $\theta_i$ must appear in the expansion exactly once. In other words, the coefficient of $\theta_1 \theta_2 \theta_3 \dots \theta_n$ is the value of the integral. There is a similar formula for the determinant, and Wick's theorem should tell you how many ways the $\theta_i$ join up with what coefficients.
To get the result the author has, you will "contract" the $\theta_i$ with each other in pairs, in all possible ways, along with the matrix elements $M_{ij}$ or $\lambda_i$ that multiply those pairs. You can crudely think of each term in this expansion as a Feynman diagram, or you can just contract them symbolically. The fact that $\det M$ factors out is like the contribution of vacuum diagrams to the path integral, which is why I suggested factoring it out early.
The standard way to get this result in physics is to introduce functional derivatives (here, ordinary derivatives):
$\int \prod_i d\theta_i \prod_j d\bar\theta_j   \exp [ \bar\theta^T (1 + M^{-1} \Lambda ) \theta] = \left.\exp\left[ \tfrac{d}{d\bar{J}_i}  (M^{-1} \Lambda)_{ij} \tfrac{d}{dJ_j}\right]  \int \prod_i d\theta_i \prod_j d\bar\theta_j   \exp [ \bar \theta^T \theta + \bar J^T \theta + J^T \bar \theta ] \right|_{J=0}$,
Now the integral is trivial (it becomes something like $e^{-\sum \bar J^T J}$), so each derivative brings down one $J_i$ or $\bar J_i$, and to have something surviving after $J=0$, you must act the derivatives in pairs. Since the exponential contains derivatives, the result will not be an exponential but a nasty sum. I suspect that $\Delta_{i_l i_k}$ is the adjugate, but I'm not sure. The best thing is to try the 3x3 case and see.
