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It is well known that given a $n \times n$ matrix $A$, it holds that

$$ \det(I + \varepsilon A)= 1 + \varepsilon \operatorname{tr}(A) + O(\varepsilon ^2).$$

I would need a full representation of $ \det(I + \varepsilon A)$, or at least some bounds explicitly depending on $A$. The reason is that I have to deal with a matrix $A$ which will depend on $\varepsilon$, in particular its entries will have to tend to infinity as $\varepsilon \to 0$ while $\operatorname{tr}(A)$ stays bounded. The problem is that, a priori, higher order terms could mess up the estimate, so I need a more precise representation. A similar question was asked here https://math.stackexchange.com/questions/1174639/series-expansion-of-the-determinant-for-a-matrix-near-the-identity, but without a source. According to the answerer, such formulas are well known by representation theorists (that's the reason of the tag), but he couldn't remember a reference.

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    $\begingroup$ If you write $I+\varepsilon A = \exp(B)$ then $\det(I+\varepsilon A) = \exp(\mathrm{Tr}(B))$. So just Taylor-expand $\log(I+\varepsilon A)$. $\endgroup$
    – Ofir Gorodetsky
    Commented Oct 23, 2023 at 20:11
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    $\begingroup$ a nice formula $${\displaystyle \det(I+A)=\sum _{k=0}^{\infty }{\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}\operatorname {tr} \left(A^{j}\right)\right)^{k}\,,}$$ is here en.wikipedia.org/wiki/Determinant#Trace. The condition is that the absolute value of every eigenvalue of A is less than 1. $\endgroup$
    – Thomas Kojar
    Commented Oct 23, 2023 at 20:28
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    $\begingroup$ $\det(I + \epsilon A) = \epsilon^{n}\det(\epsilon^{-1}I+A) = \epsilon^{n} \chi_{-A}(\epsilon^{-1})$, so we get a polynomial with degree at most $n$. $\endgroup$
    – Christophe Leuridan
    Commented Oct 23, 2023 at 20:49
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    $\begingroup$ I do not think anything useful can be obtained in such generality. E.g., suppose that the diagonal entries of $A$ are zero and the off-diagonal entries of $A$ are $\asymp1/\varepsilon$. $\endgroup$
    – Iosif Pinelis
    Commented Oct 23, 2023 at 21:00
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    $\begingroup$ Not sure if it's of help (?) but there are some interesting historical comments at this question of mine in MSE 277124 about such determinants. $\endgroup$
    – Benjamin Dickman
    Commented Oct 23, 2023 at 21:01

1 Answer 1

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$\DeclareMathOperator{\tr}{tr}$ The exact expansion is $$\det(1+tA) = \sum t^i \sigma_i(\lambda_1, \ldots, \lambda_n),$$ where $\lambda_i$ are the eigenvalues of $A$ and $\sigma_i$ is the $i$-th elementary symmetric polynomial, e.g. $\sigma_0=1, \sigma_1=\sum \lambda_i, \sigma_2=\sum _{i\ne j}\lambda_i \lambda_j, \ldots$ These are the same coefficients as in the characteristic polynomial of $A$, but in reverse order (perhaps up to sign)

As pointed out in the comments, there are ways to get these coefficients from traces of powers of your matrix, if these are any easier to come by than the eigenvalues themselves; this is because $\tr(A^k)=\sum \lambda_i^k$ forms another basis (at least rationally) of the ring of symmetric polynomials in the $\lambda_i$, so one needs only to determine the base change coefficients between this basis and the one consisting of the $\sigma_i$.

(Newton's identitites are one explicit way to do this, there may be better)

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  • $\begingroup$ I think this might settle it. $\sigma_1$ is the trace, while for the other $\sigma_i$s, I can bound the eigenvalues in terms of the $L^\infty$ norm of the matrix (Gershgoring theorem if I am not mistaken). If I make the entries of $A$ scale like, say, $\varepsilon ^{-1/2}$, the higher order terms in the sum scale as $\varepsilon$ or as $o (\varepsilon)$. $\endgroup$
    – tommy1996q
    Commented Oct 24, 2023 at 9:34
  • $\begingroup$ A question: the formula you provide holds in full generality, right? No hypothesis on triangulability of the matrix or something like that? $\endgroup$
    – tommy1996q
    Commented Oct 26, 2023 at 15:20
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    $\begingroup$ Yes, this doesn't require anything. The matrix should be square, otherwise this works in full generality, even over arbitrary coefficient fields. The proof is just multiplying $1+tA$ by $t^{-1}$ and noting that this matrix's determinant is, by definition, the characteristic polynomial of $A$ evaluated at $-t^{-1}$. $\endgroup$
    – Jan Nienhaus
    Commented Oct 27, 2023 at 18:57

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