Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F:A \to B$ map subsets of $\mathbb{R}^n$ with inverse $F^{-1}$.

Let $d(\cdot) = \text{det} \mathbf{D}F(\cdot)$ with $\mathbf{D}$ denoting the total derivative matrix.

Am I correct that $d \circ F^{-1} = \text{det} \mathbf{D}F(F^{-1}) = \text{det} \mathbf{D}\text{(Id)} = 1$.

I am not entirely sure of the what the definition of $d$ really means so I better ask this question. Thanks.

share|improve this question
    
For each $a\in A$, $\det(\mathbf{D}F(a))$ is the determinant of the matrix $\mathbf{D}F(a)$. For me it does not make sense to evaluate $\mathbf{D}F(F^{-1})$. –  Sigur Dec 23 '12 at 17:40
add comment

1 Answer

up vote 2 down vote accepted

No. The function $d\circ F^{-1}$ is $\mathbf DF\circ F^{-1}$, not $\mathbf D(F\circ F^{-1})$.

share|improve this answer
    
Thanks. Do you know any way to simplify $d \circ F^{-1}$. –  Lemon Dec 23 '12 at 18:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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