TITLE: Find the expansion for $\det(I+\epsilon A)$ where $\epsilon$ is small without using eigenvalue.
QUESTION [7 upvotes]: I'm taking a linear algebra course and the professor included the problem that prove
$$
\rm{det}(I+\epsilon A) = 1 + \epsilon\,\rm{tr}\,A + o(\epsilon)
$$
Since the professor hasn't covered the concept of eigenvalue, we are not recommended to mention eigenvalue in the solution. How to prove the equation by only using Laplace theorem?

REPLY [2 votes]: How this is done would depend on which characterizations of the determinant function you've seen.  One of those is
$$
\sum\left\{ \prod_{k=1}^n (\pm1)a_{k,\sigma(k)} : \sigma\text{ permutes }\{1,\ldots,n\} \right\}
$$
where the sign is $+1$ if $\sigma$ is an even permutation and $-1$ if it is odd.  When $\sigma$ is the identity permutation, the product is
$$
(1+\varepsilon a_{11})\cdots(1+\varepsilon a_{nn}) = 1 + \varepsilon(a_{11}+\cdots+a_{nn}) + \varepsilon^2(\cdots) + \varepsilon^3(\cdots)+\cdots.
$$
When $\sigma$ is any other permutation then the product is
$$
\varepsilon a_{k,\sigma(k)} \varepsilon a_{j,\sigma(j)}\cdots+\cdots,
$$
i.e. there are at least two indices $k$ for which $\sigma(k)\ne k$, so that $\varepsilon^2$ is a factor in this term.  Thus the determinant is
$$
1+\varepsilon\operatorname{tr}(A)+(\text{terms of degree $\ge2$ in $\varepsilon$}).
$$