Let $A,B$ be $n\times n$ real-symmetric matrices. We think of $B$ as a small perturbation of $A$.
I am looking for a bound on $$ \frac{\det(A + B) - \det A}{\det A}. $$ In particular one that is optimal in $n$.
A bound of the form $(1 + \varepsilon)^n$ is shown in [Ilse C. F. Ipsen and Rizwana Rehman, "Perturbation Bounds for Determinants and Characteristic Polynomials", SIAM J. MATRIX ANAL. APPL. Vol. 30, No. 2, pp. 762–776, https://doi.org/10.1137/070704770]. There the following is proved (Cor. 2.14): $$ \left\vert \frac{\det(A + B) - \det A}{\det A} \right\vert \leq \left(1 + \left\Vert A^{-1} \right\Vert \Vert B\Vert \right)^n - 1, $$ where $\Vert\cdot\Vert$ denotes the operator norm. (See also Bound on the difference of two determinants. )
Can one do better than a bound of the form $(1+\varepsilon)^n$?
Additional comments
By computing the derivative of $\log \det(A + tB)$ we get that $$ \log \det(A + B) - \log \det A = \mathrm{tr}[(A + t_0 B)^{-1}B] $$ for some $0 < t_0 < 1$. Thus, if one could get a good bound on $\exp(\mathrm{tr}[(A + t_0 B)^{-1}B]) - 1$ we would be done. I don't see any good way of doing this.
If it is of any help, the specific matrices I'm looking at also satisfy the following:
- $|\det A|, |\det(A + B)|\leq 1 + \varepsilon$ for some small $\varepsilon$
- $A_{ij} \geq 0$
- $A$ is block-diagonal
