Proving the Product of Jacobians Using the Chain Rule

[Lucy Yeats]

Homework Statement

The pair of variables (x, y) are each functions of the pair of variables (u, v) and vice versa.
Consider the Jacobians A=d(x,y)/d(u,v) and B=d(u,v)/d(x,y). Show using the chain rule that the product AB of these two matrices equals the unit matrix I.

Homework Equations

The Attempt at a Solution

I wrote out the two Jacobians and tried to multiply them but I got the following:
(dx/du)(du/dx)+(dx/dv)(dv/dx) (dx/du)(du/dy)+(dx/dv)(dv/dy)
(dy/du)(du/dx)+(dy/dv)(dv/dx) (dy/du)(du/dy)+(dy/dv)(dv/dy)

= 2 2dy/dx
2dy/dx 2

Where did I go wrong/ how do I continue this question?

 

[DivisionByZro]

Your product of A and B is correct, but I don't quite see how you got the result after that. Can you explain "= 2 2dy/dx " ?
2dy/dx 2

Perhaps you mean that those are what you get in the diagonals? I'm not sure.

 

[DivisionByZro]

I see what you were getting at. You're close enough to the solution, re-work the part where you simplified the partial derivatives. Check your differentials on the off-diagonal.

Edit: I removed the part about a constant multiplying the identity. Indeed B should be the inverse mapping that takes (u,v) back to (x,y), or the other way if you wish.

 

[Lucy Yeats]

Sorry, I wasn't sure how to format matrices so it's a bit unclear.
For the first row, first column: (dx/du)(du/dx)+(dx/dv)(dv/dx)=1+1=2
First row, second column: (dx/du)(du/dy)+(dx/dv)(dv/dy)= (dx/dy)+(dx/dy)=2(dx/dy) as the u's and v's cancel out.
Second row, first column:(dy/du)(du/dx)+(dy/dv)(dv/dx)=(dy/dx)+(dy/dx)=2(dy/dx) as the u's and v's cancel out.
Second row, second column:(dy/du)(du/dy)+(dy/dv)(dv/dy)=1+1=2

I'm not sure where I've made a mistake...

Thanks so much for helping! :-)

 

[Lucy Yeats]

I understand the terms on the main diagonal now, by manipulating the total derivative, but I'm still confused about the other two terms.

Any help would be great.

 

[Deveno]

if y isn't a function of x at all, and x isn't a function of y at all, what can you say about ∂y/∂x and ∂x/∂y? think of the simplest example:

x(u,v) = u
y(u,v) = v

(so that u(x,y) = x, v(x,y) = y).

 

[Lucy Yeats]

If y and x don't depend on each other, would the partial derivatives be zero?