Determinant of Matrix Component

[member 428835]

Homework Statement

Show $$\frac{\partial \det(A)}{\partial A_{pq}} = \frac{1}{2}\epsilon_{pjk}\epsilon_{qmn}A_{jm}A_{kn}$$

Homework Equations

##\det(A)=\epsilon_{ijk}A_{1i}A_{2j}A_{3k}##

The Attempt at a Solution

$$\frac{\partial \det(A)}{\partial A_{pq}}=\frac{\partial}{\partial A_{pq}}\epsilon_{ijk}A_{1i}A_{2j}A_{3k}\\
=\epsilon_{ijk}\frac{\partial}{\partial A_{pq}}A_{1i}A_{2j}A_{3k}$$
but I'm now stuck. I feel like one of the ##A## components on the RHS must go to 1 and the rest would be constant, leaving some sort of ##\epsilon A_{yy} A_{xx}## behind. I'm thinking along the lines of ##\partial A_{ij}/\partial A_{ik} = \delta_{jk}##. Any ideas?

 

[PeroK]

Any ideas?

I'd be tempted to move ##A_{pq}## to the top-left of the matrix.

 

[member 428835]

I'd be tempted to move ##A_{pq}## to the top-left of the matrix.

Are you suggesting take
$$\frac{\partial \det(A)}{\partial A_{11}}=\frac{\partial}{\partial A_{11}}\epsilon_{ijk}A_{1i}A_{2j}A_{3k}\\
=\epsilon_{ijk}A_{2j}A_{3k}\frac{\partial}{\partial A_{11}}A_{1i}\\
=\epsilon_{1jk}A_{2j}A_{3k}$$

 

[PeroK]

Are you suggesting take
$$\frac{\partial \det(A)}{\partial A_{11}}=\frac{\partial}{\partial A_{11}}\epsilon_{ijk}A_{1i}A_{2j}A_{3k}\\
=\epsilon_{ijk}A_{2j}A_{3k}\frac{\partial}{\partial A_{11}}A_{1i}\\
=\epsilon_{1jk}A_{2j}A_{3k}$$

One way to do it is just to do it for all 9 elements. Not very elegant.

Otherwise, you need to get a general expression for the terms that have ##A_{pq}## in them. It might be easier to rearrange the matrix, getting ##A_{pq}## to the top left corner.

 

[member 428835]

One way to do it is just to do it for all 9 elements. Not very elegant.

Agreed!

Otherwise, you need to get a general expression for the terms that have ##A_{pq}## in them. It might be easier to rearrange the matrix, getting ##A_{pq}## to the top left corner.

What do you mean "general expression"? Also, you suggest getting ##A_{pq}## to the top left, doesn't this mean making ##A_{pq}=A_{11}##?

 

[PeroK]

Agreed!

What do you mean "general expression"? Also, you suggest getting ##A_{pq}## to the top left, doesn't this mean making ##A_{pq}=A_{11}##?

It means using row and column operations. Alternatively, express the determinant in an equivalent form for expansion via the ##p##th row.

 

[member 428835]

It means using row and column operations. Alternatively, express the determinant in an equivalent form for expansion via the ##p##th row.

Since the determinant is a scalar, I'm unsure how to express it for the ##p##th row. Could you elaborate?

 

[PeroK]

Since the determinant is a scalar, I'm unsure how to express it for the ##p##th row. Could you elaborate?

You can evaluate a determinant using any row as your starting row.

But, in fact, if you permute the rows to get the ##p##th row at the top, with row ##p+1## second and row ##p+2## third (where these are numbers modulo 3), I think it comes out easily enough. Then permute the columns to get the colums in order ##q, q+1, q+2##.

Then, if you do the same trick with the Levi-Civita, I think the equation should drop out for general ##p, q##.

 

[PeroK]

PS by the same trick, I mean that:

##\epsilon_{pjk} = 1## when ##j = p+1## and ##k = p+2##
##\epsilon_{pjk} = -1## when ##j = p+2## and ##k = p+1##

And is ##0## otherwise.

 

[member 428835]

You can evaluate a determinant using any row as your starting row.

I agree.

But, in fact, if you permute the rows to get the ##p##th row at the top, with row ##p+1## second and row ##p+2## third (where these are numbers modulo 3), I think it comes out easily enough. Then permute the columns to get the colums in order ##q, q+1, q+2##.

I don't think I'm understanding what you're saying. Could you illustrate, perhaps with a different problem? I just don't know the technique you suggest.

PS by the same trick, I mean that:

##\epsilon_{pjk} = 1## when ##j = p+1## and ##k = p+2##
##\epsilon_{pjk} = -1## when ##j = p+2## and ##k = p+1##

And is ##0## otherwise.

I agree.

 

[PeroK]

Suppose we take the case ##p, q = 1, 1##:

##det(A) = A_{11}(A_{22}A_{33} - A_{23}A_{32}) + \dots##

Where the rest of the terms do not involve ##A_{11}##, hence you can calculate the required partial derivative.

You can also check fairly easily that:

##\frac{1}{2}\epsilon_{1jk}\epsilon_{1mn}A_{jm}A_{kn} = A_{22}A_{33} - A_{23}A_{32}##

Now, by either two row operations or no row operations and two column operations or no column operations, you permute ##A_{pq}## to the top left of your determinant. You need to check this for ##p, q = 1, 2, 3##. And, this gives

##det(A) = A_{pq}(A_{p+1, q+1}A_{p+2, q+2} - A_{p+1, q+2}A_{p+2, q+1}) + \dots##

Where we are using modulo 3 here. And that's just about it. All you need to do is manipulate the expression involving the Levi_Civita coefficients using a similar modulo 3 approach.

 

[member 428835]

Thanks!