- #1
vst98
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Homework Statement
Prove the chain rule for Jacobi determinants
[itex] \frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)}=\frac{d(f,g)}{d(x,y)}[/itex]
Homework Equations
Definition of Jacobi determinant
[itex] \frac{d(f,g)}{d(u,v)} = \frac{d(f,g)}{d(u,v)} = det \begin{bmatrix}
\frac{df}{du}&\frac{df}{dv} \\
\frac{dg}{du}&\frac{dg}{dv}
\end{bmatrix} [/itex]
The determinant of a matrix product of square matrices equals the product of their determinants:
det(AB)=det(A)det(B)
The Attempt at a Solution
[itex]\frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)} = det \begin{bmatrix}
\frac{df}{du}&\frac{df}{dv} \\
\frac{dg}{du}&\frac{dg}{dv}
\end{bmatrix} *
det \begin{bmatrix}
\frac{du}{dx}&\frac{du}{dy} \\
\frac{dv}{dx}&\frac{dv}{dy}
\end{bmatrix} = det (\begin{bmatrix}
\frac{df}{du}&\frac{df}{dv} \\
\frac{dg}{du}&\frac{dg}{dv}
\end{bmatrix} *
\begin{bmatrix}
\frac{du}{dx}&\frac{du}{dy} \\
\frac{dv}{dx}&\frac{dv}{dy}
\end{bmatrix}) = det\begin{bmatrix}
\frac{df}{du}\frac{du}{dx} + \frac{df}{dv}\frac{dv}{dx}&\frac{df}{du}\frac{du}{dy} + \frac{df}{dv}\frac{dv}{dy} \\
\frac{dg}{du}\frac{du}{dx} + \frac{dg}{dv}\frac{dv}{dx} & \frac{dg}{du}\frac{du}{dy} + \frac{dg}{dv}\frac{dv}{dy}
\end{bmatrix} = det \begin{bmatrix}
2\frac{df}{dx} & 2\frac{df}{dy} \\
2\frac{dg}{dx}&2\frac{dg}{dy}
\end{bmatrix} = 2*\frac{d(f,g)}{d(x,y)}[/itex]
which is obviously wrong but I don't see where is my error.
Is it wrong to cancel out those cross terms after multiplying matrices, but what else can I do ?