chill_factor said:Homework Statement
L = lambda.
Prove: d(A^-1)/dL = -(A^-1)(dA/dL)(A^-1)
Homework Equations
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The Attempt at a Solution
I did this as an analogy with function of numbers, but don't know how to extend this to matricies. for example:
lets say A = f(L)
d(f(L)^-1)/dL = - (f(L)^-2*d(f(L))/dL = -(A^-1)*dA/dL*(A^-1)
But what is the matrix form?
Ray Vickson said:I'll use x instead of L, and let B(x) = Inv(A(x)); thus, A(x)*B(x) = I (identity matrix). Take the derivative.
RGV
The notation d(A^-1)/dL represents the derivative of the inverse of matrix A with respect to the variable L. In other words, it measures how the inverse of A changes as the variable L changes.
A^-1 being a function of L means that the elements of the inverse matrix A^-1 are dependent on the value of the variable L. This usually happens when the original matrix A is also a function of L.
The expression d(A^-1)/dL is related to dA/dL through the chain rule of differentiation. Specifically, it can be written as d(A^-1)/dL = -(A^-1)(dA/dL)(A^-1), where dA/dL represents the derivative of matrix A with respect to L.
The negative sign is present because the derivative of the inverse of a function is equal to the negative of the inverse of the function multiplied by the derivative of the function itself. In this case, the inverse function is A^-1 and the function is A, hence the negative sign.
No, the expression d(A^-1)/dL = -(A^-1)(dA/dL)(A^-1) cannot be simplified further as it represents the derivative of a matrix with respect to a variable. However, the individual elements of the matrix can be simplified if the original matrix A is known.