Derivative of determinant wrt matrix

[bakav]

Hi there,
I want to derive the derivative of the det(A+O'XO) with respect to X, where A, O', O and X are all matrix.
Any suggestions,
Thanks
Baska

 

[Simon_Tyler]

First note that
det(A+O'XO) = exp(tr(log(A+O'XO)))

Then define the matrix partial derivative d_X such that
d_X tr(Xⁿ) = n X^n-1
In terms of components, this means
(d_X f(X))_ij = d_Xji f(X)
where f(X) is an arbitrary scalar function of X or some component of a vector/matrix/tensor function of X.

Then we take the derivative
d_X det(A+O'XO)
= d_X exp(tr(log(A+O'XO)))
= exp(tr(log(A+O'XO))) d_Xtr(log(A+O'XO))
= det(A+O'XO) tr(d_Xlog(A+O'XO))
= det(A+O'XO) tr((A+O'XO)^-1d_X(A+O'XO))
= det(A+O'XO) O(A+O'XO)^-1O'
= O adj(A+O'XO) O'

where adj(A) = det(A) A^-1 is the http://en.wikipedia.org/wiki/Adjugate" of A. (Note that in older texts you see it called the adjoint - but that gets confusing with other adjoints).

Although the above definition goes via log(A+O'XO), the result does not require its existence. By looking at the definition of determinant, you can find it's derivative directly in terms of the adjugate. http://en.wikipedia.org/wiki/Determinant#Derivative
Note the transposition of the indices in their formula - which justifies my definition of matrix partial derivative.

Also see
http://en.wikipedia.org/wiki/Matrix_calculus#Derivative_of_matrix_determinant

 

[bakav]

Thanks man
I also tried to use the general formula as:

d(det(Y))/dX=dtr(QY)/dX where Q is the det(Y)inv(Y) and apparently Y is a function of X.
I got the same derivation as you did.
Thanks a lot for your great help.

 

[bakav]

thanks dude appreciate it.
Can you guide me that how you got, in your derivation, from
= exp(tr(log(A+O'XO))) dXtr(log(A+O'XO)
= det(A+O'XO) tr(d_Xlog(A+O'XO))
= det(A+O'XO) tr((A+O'XO)^-1d_X(A+O'XO))
= det(A+O'XO) O(A+O'XO)^-1O'
My question addresses how did you bring in the partial derivative dX to the trace function and then how you got rid of the trace in the last equation.