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Determinant about identity

The following pops up in many places, one example being the proof of Liouville’s theorem in Hamiltonian mechanics. There, it arises from the Jacobian associated with an infinitesimal time evolution.

For notational convenience, I will denote equality up to linear order in \( \epsilon \) with \( \approx \).

Claim. Given any matrix \( M \), one has

$$ \mathrm{det}(I + \epsilon M) = 1 + \epsilon \mathrm{Tr}(M) + \mathcal{O}(\epsilon^2) \approx 1 + \epsilon \mathrm{Tr}(M), $$
where \( I \) is the identity matrix.

Proof 1. There are many ways to proceed, the purest is to start with the following definition of the determinant:

$$ \mathrm{det}(M) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi) \prod_{i=1}^n M_{i\pi(i)}, $$
where \( S_n \) is the symmetry group of order \( n \), and \(\mathrm{sgn}(\pi) \) equals \( +1\) for even and \( -1 \) for odd permutations.

It follows that

$$ \mathrm{det}(I + \epsilon M) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi) \prod_{i=1}^n (\delta_{i\pi(i)} + \epsilon M_{i\pi(i)}) $$

Here comes the clever bit: suppose \(\pi\) is not the identity element in \(S_n\). Then, there exists at least two terms with \(i \neq \pi(i)\), which implies the product has two factors of \(\epsilon\) as the corresponding \(\delta_{i\pi(i)}\) evaluate to zero. Since we only work to linear order in \(\epsilon\), we can ignore all permutations except for the identity! The result follows:

$$ \mathrm{det}(I + \epsilon M) \approx \prod_{i=1}^n (1 + \epsilon M_{ii}) \approx 1 + \epsilon \sum_{i=1}^n M_{ii} = 1 + \epsilon \mathrm{Tr}(M). $$

Proof 2. An easier proof is to work in terms of the eigenvalues, in terms of which the determinant and trace take the forms:

$$ \mathrm{det}(M) = \prod_i \lambda_i, \qquad \mathrm{Tr}(M) = \sum_i \lambda_i. $$
The result directly follows:
$$ \mathrm{det}(1 + \epsilon M) = \prod_i (1 + \epsilon \lambda_i) \approx 1 + \epsilon \sum_i \lambda_i = 1 + \epsilon \mathrm{Tr}(M). $$