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Dusan Pokorny and Jan Rataj have just posted a paper ( 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. $(d-k)f + kg, \ 0\le k \le d$).

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Are there conditions on A and B? I have problems with it when d is even and A=B. Gerhard "Ask Me About System Design" Paseman, 2012.09.13 –  Gerhard Paseman Sep 13 '12 at 17:13
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 Sep 13 '12 at 18:39
(And yes, this is well-known. I can't give a name, but it's a basic fact from the theory of finite differences.) –  darij grinberg Sep 13 '12 at 18:40
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 D. Elkies Sep 13 '12 at 18:41
[and darij grinberg wrote much the same thing a minute or two earlier.] –  Noam D. Elkies Sep 13 '12 at 18:42

2 Answers 2

I have not seen precisely this identity before, but it seems to be a variation on Chapman's identity [R. Chapman, Amer. Math. Monthly 109(7) (2002), 664–666], see also arXiv:math/0612464,

$\sum_{x_1,x_2,\ldots x_N=0}^{1}(-1)^{x_1+x_2+\cdots x_N} \;\text{det}\;(x_1 A_1+x_2 A_2 +\cdots x_N A_N)=0,$

valid for any set $A_1,A_2,\ldots A_N$ of $d\times d$ matrices, with $N\geq d+1$.

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But I don't see that the identity under discussion follows from Chapman's identity. –  Joe Fu Sep 13 '12 at 21:16
(which is too bad, since I'm new here and could use the 2 reputation points) –  Joe Fu Sep 13 '12 at 21:18
indeed, with "a variation on" I meant that it is not "a special case of", but both equate an alternating sum of determinants of sums of matrices to zero. –  Carlo Beenakker Sep 14 '12 at 5:32
up vote 4 down vote accepted

See the comments of Elkies and Grinberg.

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