Second Derivative of Determinant of Matrix?

[brydustin]

Hi all...


I've read on wikipedia (facepalm) that the first derivative of a determinant is

del(det(A))/del(A_ij) = det(A)*(inv(A))_j,i

If we go to find the second derivative (applying power rule), we get:

del^2(A) / (del(A)_pq) (del (A)_ij) = {del(det(A))/del(A_pq)}*(inv(A))_j,i + det(A)*{del(inv(A)_j,i) / del(A_pq)}

I have no clue how to calculate the derivative of the inverse of a matrix with respect to changing the values in the original matrix:
I.E. del(inv(A)_j,i) / del(A_pq)

Also... would be nice if someone could prove the first statement for the first derivative of the determinant.

Thanks!

 

[fresh_42]

You can use the closed formula $\det A = \sum_{\sigma \in \operatorname{Sym}(n)} \operatorname{sgn}(\sigma) \prod_{k=1}^n A_{k\sigma(k)}$ and calculate.

 

[Cryo]

I have written about the first derivative of the determinant here

https://www.physicsforums.com/threa...-cofactor-and-determinant.970419/post-6165630
given a matrix $A$ that depends on some variable $x$: $A_{ij}=A_{ij}\left(x\right)$, the derivative of its determinant ($A=\mbox{det}\left(A\right)$) is:

$\partial_x A = A \left(A^{-1} \right)_{ji} \partial_x A_{ij}$

if $x\to A_{sk}$ then $\partial_{x} A_{ij} \to \delta_{si}\delta_{kj}$ so:$\partial_{A_{sk}} A = A \left(A^{-1} \right)_{ks}$

 

[WWGD]

Hi all...I've read on wikipedia (facepalm) that the first derivative of a determinant is

del(det(A))/del(A_ij) = det(A)*(inv(A))_j,i

If we go to find the second derivative (applying power rule), we get:

del^2(A) / (del(A)_pq) (del (A)_ij) = {del(det(A))/del(A_pq)}*(inv(A))_j,i + det(A)*{del(inv(A)_j,i) / del(A_pq)}

I have no clue how to calculate the derivative of the inverse of a matrix with respect to changing the values in the original matrix:
I.E. del(inv(A)_j,i) / del(A_pq)

Also... would be nice if someone could prove the first statement for the first derivative of the determinant.

Thanks!

Can't you use $Det(AA^{-1})=Id$ to differentiate?

 

[fresh_42]

Can't you use $Det(AA^{-1})=Id$ to differentiate?

... and shouldn't the trace be $D_{Id}\det A\,$?

 

[Mark44]

Can't you use $Det(AA^{-1})=Id$ to differentiate?

Shouldn't Id in the above equation be just 1? $Det(I) = 1$.

 

[WWGD]

Shouldn't Id in the above equation be just 1? $Det(I) = 1$.

Ah,yes,Duh myself.