Let $A$ and $B$ be two real, $n\times n$ matrices. Using Hadamard's inequality, it is not hard to show that $$ \left|\det A - \det B \right| \leq \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. $$ Where $\|A\|_2=\sqrt{\sum_{i,j}a_{ij}^2}$. From this, I can derive a sup bound, for example $$ \left|\det A - \det B \right| \leq n^{n+1} \|A-B\|_{\infty} \max (\|A\|_{\infty}^{n-1},\|B\|_{\infty}^{n-1}). $$ Where $\|A\|_\infty=\sup_{i,j}|a_{ij}|$.

The constant $n^{n+1}$ is not the best bound possible : any reference (or proof) for a better (or the best) one? I show below that one can obtain $n^2(n-1)^{n-1}$, but that isn't much better. I just tried $10^5$ random matrices on Maple and obtained a maximal constant (much) smaller than one : this is not a proof, but it looks like there is room for improvement nevertheless.

Just for completeness (and in case someone sees a factor I missed), to get the first bound, writing $A=[A_1,\ldots,A_n]$ in terms of its column vectors, an expansion shows \begin{eqnarray*} \det A &=& \det (A_1 -B_1,A_2,\ldots,A_n) + \det (B_1,A_2,\ldots,A_n) \\ &=& \sum_{j=1}^n \det (B_1,\ldots, B_{j-1}, A_j -B_j,A_{j+1},\ldots,A_n) \\ && + \det B, \end{eqnarray*} Thus by Hadamard's inequality, \begin{eqnarray*} \det A -\det B &\leq& \sum_{j=1}^n \prod_{i=1}^{j-1} \|B_i\|_{2}\prod_{i=j+1}^{n} \|A_i\|_{2} \|A_j-B_j\|_{2} \\ &\leq& \|A-B\|_{2} \sum_{j=1}^n \|B\|^{j-1}_{2}\|A\|^{n-j}_{2} \\ &=& \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. \end{eqnarray*}

The second bound is just that $x^n -y^n\leq n \max(|x|^{n-1},|y|^{n-1}) |x-y|$ and $\|A\|_2 \leq n\|A\|_\infty$.

Another approach is calculus, namely, to write that $\det B - \det A = f(1)-f(0)$, with $f(t)=\det(A + t(B-A))$.

By the mean value theorem $f(1)-f(0)\leq \max |f^\prime (t)|$.

We can compute that $$f^\prime(t) = {\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)$$ (if I did not mess up, using the formula for the differential of a determinant, where Cofm means the matrix of Cofactors).

Then, it should deliver something better, if there is a nice way to bound it. The simplest thing is to use Cauchy-Schwarz, namely $$ \left|{\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)\right|\leq \|B-A\|_{2} \|{\rm Cofm}(A +t(B-A))\|_{2} $$ and then, by lack of a better idea, $$ \|{\rm Cofm}(A +t(B-A))\|_{2}\leq n \max_{ij} |{\rm Cof}_{i,j}(A +t(B-A))|, $$ and brutally, $|{\rm Cof}_{i,j}(A +t(B-A))|\leq ((n-1) \max(\|A\|_\infty,\|B\|_\infty))^{n-1}$ gives a slightly better constant, namely $$ n^2(n-1)^{n-1} <n^{n+1} $$ but that still seems a very rough way to bound a determinant, as it is never sharp, since to attain this bound all coefficients should be equal, and therefore the cofactor would be zero. They are of the same order in $n$, and I suspect this order is wrong.


1 Answer 1


Since the most of related results seem to be published in articles devoted to the perturbation of determinant, let me rewrite your bound $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,B\,\right)\,\big\rvert \le n^{n+1} \left\|\,A-B\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,B\,\right\|^{n-1}_\infty \big\}\tag{1a}\label{1a} $$ under assumption $\,B = A + E$: $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le n^{n+1} \left\|\,E\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,A+E\,\right\|^{n-1}_\infty \big\}\tag{1b}\label{1b} $$

The best bound I could find is presented in this paper:

Absolute perturbation bounds (Theorem 2.12, p.768, [1]):

Let $A$ and $E$ be $n \times n$ complex matrices. Then $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le n \,\left\|\,E\,\right\|_2 \max \big\{ \left\|\,A\,\right\|_2, \,\left\|\,A+E\,\right\|_2 \big\}^{n-1} \tag{2}\label{2} $$

Relative perturbation bounds (Corollary 2.14, p.770, [1]):

Let $A$ and $E$ be $n \times n$ complex matrices. If $A$ is nonsingular, then $$ \frac{ \big\vert \, \det \left(\,A+E\,\right) - \det\left(\,A\,\right)\,\big\rvert }{ \big\vert\,\det \left(\,A\,\right)\,\big\rvert } \le \Big( \big\|\,A^{-1}\,\big\|_2 \big\|\,E\,\big\|_2\Big)^{n}-1 = \left(\kappa\,\frac{\left\|\,E\,\right\|_2}{\left\|\,A\,\right\|_2}\right)^{n}-1, \tag{3}\label{3} $$ where $ \kappa \equiv \big\|\,A\,\big\|_2 \,\big\|\,A^{-1}\,\big\|_2$.

The absolute bound result can be rewritten in terms of singular values of $A$.

Absolute perturbation bounds (Corollary 2.7, p.767, [1]): Let $A$ and $E$ be $n \times n$ complex matrices. Then $$\big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le\sum_{i=1}^{n} s_{n-i} \left\|E\right\|^i_2.\tag{2a}\label{2a}$$ If $\operatorname{rank}\left(A\right) = r$ for some $1 ≤ r ≤ n − 1$, then $$\big\vert \det \left(\,A+E\,\right)\big\rvert \le\left\|E\right\|^{n-r}_2 \sum_{i=1}^{r} s_{r-i} \left\|E\right\|^i_2,$$ where the $s_j$ are elementary symmetric functions in the $r$ largest singular values of $A$, $1 ≤ j ≤ r$. The bounds hold with equality for $E = \varepsilon UV_*$ with $ε > 0$, where $ A = UΣV_*$ is a SVD of $A$.

Let us compare, for example, $\eqref{1b}$ and $\eqref{2}$. Since $$ \frac{1}{\sqrt n}\left\|\,E\,\right\|_\infty \le \left\|\,E\,\right\|_2 \le \sqrt n\left\|\,E\,\right\|_\infty \implies \left\|\,E\,\right\|_\infty \ge n^{-\frac{1}{2}} \left\|\,E\,\right\|_2 , $$ we have $$ \begin{aligned} \eqref{1b} & = n^{n+1} \left\|\,E\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,A+E\,\right\|^{n-1}_\infty \big\} \ge \\ & \ge n^{n} \left( n^{-\frac{1}{2}}\left\|\,E\,\right\|_2\right) \max\big\{\left\|\,A\,\right\|_\infty,\,\left\|\,A+E\,\right\|_\infty\big\}^{n-1} = \\ & = n^{n+\frac{1}{2}} \left\|\,E\,\right\|_2 \max\big\{\left\|\,A\,\right\|_\infty,\,\left\|\,A+E\,\right\|_\infty\big\}^{n-1} \ge \\ & \ge n^{\frac{n}{2}+1} \left\|\,E\,\right\|_2 \max\Big\{\!\!\left( n^{-\frac{1}{2}}\left\|\,A\,\right\|_2\right), \, \left( n^{-\frac{1}{2}}\left\|\,A+E\,\right\|_2\right)\!\!\Big\}^{n-1} = \\ & = n^{\frac{n+3}{2}}\left\|\,E\,\right\|_2 n^{\frac{1-n}{2}} \max\Big\{\left\|\,A\,\right\|_2,\,\left\|\,A+E\,\right\|_2 \Big\}^{n-1} = \\ & = n^{2} \left\|\,E\,\right\|_2 \max\Big\{\left\|\,A\,\right\|_2,\,\left\|\,A+E\,\right\|_2 \Big\}^{n-1} \ge \\ & \ge n \left\|\,E\,\right\|_2 \max \big\{ \left\|\,A\,\right\|_2, \,\left\|\,A+E\,\right\|_2 \big\}^{n-1} = \eqref{2} \end{aligned} $$

Thus, we conclude that the estimate $\eqref{2}$ is sharper than $\eqref{1b}$.


  1. Ipsen, Ilse C. F., and Rizwana Rehman. "Perturbation Bounds for Determinants and Characteristic Polynomials." SIAM. J. Matrix Anal. & Appl. SIAM Journal on Matrix Analysis and Applications 30.2 (2008): 762-76. Web. 7 Aug. 2015.
  • 1
    $\begingroup$ Sorry for not giving the bounty right away (I had gone on vacation), luckily the site did it for me. Thank you. A little disagreement with your bounds at the beginning of the comparison (but it does not matter). Since the size of $E$ is $n^2$, it should be $n$, not $\sqrt{n}$. $\endgroup$
    – user145413
    Aug 18, 2015 at 20:13
  • $\begingroup$ ...which means (2) is better than (1b), but just by $\geq$, without a factor. And it isn't so clearly better than (1b) with $n^2(n-1)^{n-1}$ for large $n$. $\endgroup$
    – user145413
    Nov 14, 2017 at 11:45
  • $\begingroup$ @username have you found a better bound by now? $\endgroup$
    – Marcel
    Jul 1, 2019 at 9:23
  • $\begingroup$ @marcel I am happy with the bounds given in the answer. The paper quoted, and subsequent articles give lots of variants, which work for me. $\endgroup$
    – user145413
    Aug 2, 2019 at 14:18

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