Re: Derivative of the Determinant Function
- From: Ronald Bruck <bruck@xxxxxxxxxxxx>
- Date: Sat, 06 May 2006 17:58:08 -0700
In article <1146959828.623730.186290@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Chip Eastham <hardmath@xxxxxxxxx> wrote:
Corey wrote:
Hi there. If I were to compute the derivative of the determinant
function as a function of the linear space of n x n matrices, how would
I go about doing this? My initial thought was to treat this as a Hodge
duality problem, but I'm not really sure where to begin if I take this
approach. I was just wondering if anyone could offer me some input,
Hi, Corey:
Mr. Petry points out that there is room for confusion in understanding
the question. In one sense the determinant is taken as a scalar
function of several (scalar) variables, and then one can ask for the
partial derivatives wrt to individual matrix entries, ie. the signed
cofactors.
Another sense is to treat the determinant, as a scalar function of
a single "vector valued" argument, one that ranges over the linear
space of nxn matrices. One can ask for "directional derivatives"
specifying a "point" (matrix A) at which the determinant is to be
evaluated and a "directional increment" (matrix H), corresponding
to which the directional derivative is expressed by generalizing
the usual limit of a difference quotient:
LIMIT ( det(A+eH) - det(A) ) / eH
e-->0
Of course we are going to replace division by H by multiplication
by its inverse, but in doing so we point out that the numerator is
always a scalar and hence commutes universally with H^-1, so
this gives a well-defined approach.
Having laid out these two alternative interpretations of your
"take the derivative" question, I leave it to you to work out the
close relationship between the partial derivatives in the former
and the directional derivative in the latter, which both may be
stated in reference to the adjoint of A.
In any case, the partial derivatives are trivial to compute by
expanding by minors. If A(i|j) denotes the matrix with the i-th column
and j-th row deleted, then
det A = \sum_j (-1)^(i+j) a_{ij} det A(i|j),
so for a fixed i and j, none of the det A(i|k) involve a_{ij} and so
\partial det A/\partial a_{ij} = (-1)^(i+j) det A(i|j).
You can also see this by using the definition det A = \sum_\sigma
sgn(\sigma) \product a_{i,\sigma(i)}. (If that IS your definition.)
--
Ron Bruck
.
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