Derivative of determinant wrt matrix

In summary, the conversation is about deriving the derivative of the determinant of a matrix expression with respect to one of the matrices involved. One person provides a detailed explanation using the definition of the adjugate and the matrix partial derivative, while the other person asks for clarification on certain steps of the derivation.
  • #1
bakav
7
0
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
 
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  • #2
First note that
det(A+O'XO) = exp(tr(log(A+O'XO)))

Then define the matrix partial derivative dX such that
dX tr(Xn) = n Xn-1
In terms of components, this means
(dX f(X))ij = dXji 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
dX det(A+O'XO)
= dX exp(tr(log(A+O'XO)))
= exp(tr(log(A+O'XO))) dXtr(log(A+O'XO))
= det(A+O'XO) tr(dXlog(A+O'XO))
= det(A+O'XO) tr((A+O'XO)-1dX(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" [Broken] 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
 
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  • #3
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.
 
  • #4
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(dXlog(A+O'XO))
= det(A+O'XO) tr((A+O'XO)-1dX(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.
 
  • #5


Hi Baska,

Thank you for your question. The derivative of the determinant with respect to a matrix is a fundamental concept in linear algebra and can be derived using the properties of determinants and matrix calculus.

To begin, let's define the determinant of a matrix A as det(A). The derivative of the determinant with respect to a matrix can be written as d(det(A))/dA. Using the chain rule, we can rewrite this as d(det(A))/dA = d(det(A))/dA * dA/dX.

Next, we can expand the determinant of A+O'XO using the Leibniz formula, which states that the determinant of a sum of matrices is equal to the sum of the determinants of each individual matrix. This gives us det(A+O'XO) = det(A) + det(O'XO) + det(AO'XO).

Now, using the properties of determinants, we can rewrite det(AO'XO) as det(A) * det(O'XO). Substituting this back into our original equation, we have det(A+O'XO) = det(A) + det(O'XO) + det(A) * det(O'XO).

Next, let's consider the derivative of det(O'XO) with respect to X. Using the properties of matrix calculus, we can rewrite this as d(det(O'XO))/dX = det(O'XO) * d(O'XO)/dX.

Now, using the chain rule again, we can rewrite d(O'XO)/dX as d(O'XO)/dX = d(O'XO)/dO' * dO'/dX * dX/dX. Since O' is a constant matrix, d(O'XO)/dO' = X, and dO'/dX = 0. Therefore, d(O'XO)/dX = X * dX/dX = X.

Substituting this back into our original equation, we have det(A+O'XO) = det(A) + det(O'XO) + det(A) * det(O'XO) * X.

Finally, taking the derivative of both sides with respect to X, we get d(det(A+O'XO))/dX = det(O'XO) + det(A) * X. Therefore,
 

What is the derivative of a determinant with respect to a matrix?

The derivative of a determinant with respect to a matrix is a matrix of the same size, where each element is the cofactor of the corresponding element in the original matrix.

Why do we need to take the derivative of a determinant with respect to a matrix?

Taking the derivative of a determinant with respect to a matrix allows us to calculate the sensitivity of the determinant to small changes in the elements of the matrix. This is useful in optimization problems, where we need to find the maximum or minimum value of a determinant.

How is the derivative of a determinant with respect to a matrix calculated?

The derivative of a determinant with respect to a matrix is calculated using the cofactor expansion method. This involves finding the cofactors of each element in the matrix and constructing a new matrix with these cofactors as elements.

Is the derivative of a determinant with respect to a matrix always a square matrix?

Yes, the derivative of a determinant with respect to a matrix is always a square matrix of the same size as the original matrix. This is because the cofactor expansion method results in a square matrix.

Can the derivative of a determinant with respect to a matrix be used in other areas of mathematics?

Yes, the derivative of a determinant with respect to a matrix has applications in fields such as linear algebra, multivariate calculus, and optimization. It is also used in computer graphics and computer vision for calculating transformations and deformations of objects.

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