6
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ There's a typo in your formula for $G^{-1}$. $\endgroup$
    – Stephen
    Jul 14 '12 at 13:49
6
$\begingroup$

$G$ 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!

$\endgroup$
1
  • $\begingroup$ That's the point, you're right. Thanks! $\endgroup$
    – bartgol
    Jul 14 '12 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.