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!