Dusan Pokorny and Jan Rataj have just posted a paper (<http://arxiv.org/abs/1209.2305>) in which they prove the identity
$$
\det (A-B) = \frac 1{d!} \sum\_{k=0}^d (-1)^k \binom dk \det((d-k)A + kB)
$$
where $A,B$ are $d \times d$ matrices (this is the corrected version of the formula, in response to Paseman's initial incredulity).


This is so simple and beautiful that one is tempted to suspect that it is "known". Has anyone seen this before?


I should mention Pokorny-Rataj's application: if $f,g$ are (nonsmooth) convex functions on $\mathbb R^n$ , then there exists a signed measure on $\mathbb R^n$ that stands in for the integral of the determinant of the Hessian of $f -g$. In fact, there exists a closed integral current (in the sense of Federer-Fleming) in $\mathbb R^n \times \mathbb R^n$ that stands in for the graph of $\nabla(f-g)$ . Using the identity, this follows from the classical fact that this is true of any convex function (e.g.
$(d-k)f + kg, \ 0\le k \le d$).



a determinantal identity
See the comments of Elkies and Grinberg: 


*Looks like an application of the finite-difference formula to
the polynomial $P(k) := \det(dA - k(A-B))$ which has degree $d$
with leading coefficient $(-1)^d \det(A-B)$.* (Noam Elkies) 


*More generally: if $A$ is a commutative ring, and $P\in A\left[X\right]$ is a polynomial of degree $\leq d$, then $\dfrac{1}{d!}\sum\limits\_{k=0}^{d} \left(-1\right)^k \dbinom{d}{k} P\left(k\right)$ equals $\left(-1\right)^d$ times the $X^d$-coefficient of $P$.* (Darij Grinberg)


