I have a doubt on Jacobian matrices. Consider the non linear transformation

$$ \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] = \mathbf{G}\left( \left[ \begin{array}{c} \hat{x}\\ \hat{y}\\ \hat{z} \end{array}\right] \right) = \left[ \begin{array}{c} \hat{x}g(\hat{z})\\ \hat{y}g(\hat{z})\\ \hat{z} \end{array}\right] $$ whose Jacobian reads

$$ \text{J} = \left[ \begin{array}{ccc} g & 0 & \hat{x}g'\\ 0 & g & \hat{y}g'\\ 0 & 0 & 1 \end{array} \right] $$

If I invert this matrix I get

$$ \text{J}^{-1} = \left[ \begin{array}{ccc} 1/g & 0 & -\hat{x}g'/g\\ 0 & 1/g & -\hat{y}g'/g\\ 0 & 0 & 1 \end{array} \right] $$ which I thought should be the same as the Jacobian of the inverse transformation. However, solving for $\hat{x}, \hat{y}, \hat{z}$ in the definition of the transformation, I get

$$ \left[ \begin{array}{c} \hat{x}\\ \hat{y}\\ \hat{z} \end{array}\right] = \mathbf{G}^{-1}\left( \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \right) = \left[ \begin{array}{c} x/g(z)\\ y/g(z)\\ z \end{array}\right] $$ whose Jacobian now reads

$$ \text{J}^{-1} = \left[ \begin{array}{ccc} 1/g & 0 & -\hat{x}g'/g^2\\ 0 & 1/g & -\hat{y}g'/g^2\\ 0 & 0 & 1 \end{array} \right] $$ which is slightly different. My question is: which one is the correct Jacobian for the inverse? Weren't they supposed to be the same? If so, where's my mistake?

Thank you in advance!

  • 1
    $\begingroup$ There's a typo in your formula for $G^{-1}$. $\endgroup$
    – Stephen
    Jul 14 '12 at 13:49

$G$ maps a point $p$ to $G(p)$. The Jacobian maps a tangent vector at $p$ to one at $G(p)$. The inverse is the Jacobian for $G^{-1}$ at $G(p)$. So, in the second formula you should substitute $x g(z)$ for $x$, $yg(z)$ for $y$, and $z$ for $z$ to recover the first. All consistent (modulo the typo I mentioned in the comment), well done!

  • $\begingroup$ That's the point, you're right. Thanks! $\endgroup$
    – bartgol
    Jul 14 '12 at 14:06

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