# derivations – What’s the relation between Lipschitz ceaseless and the determinant of Jacobian matrix? Answer

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derivations – What’s the relation between Lipschitz ceaseless and the determinant of Jacobian matrix?

The absolute worth of the determinant of the Jacobian matrix of the transformation $$f$$ exhibits how mighty $$f$$ adjustments the $$n$$-volume, domestically. The Lipschitz ceaseless of $$f$$ exhibits how mighty $$f$$ adjustments the lengths, on the most.

Nik Weaver talked about that the Lipschitz quantity is all the time not less than the $$n$$th root of absolutely the worth of the determinant of the Jacobian fairly believable.

The goal right here is to supply a proof of this assertion. Indeed, approximating domestically $$f$$ by an affine transformation, we could occupy that $$f$$ is a linear transformation given by an $$ntimes n$$ actual matrix $$A$$. Then the talked about assertion can breathe rewritten as
$$|det A|^{1/n}overset{textual content{(?)}}le|A|, tag{1}$$
the place $$|A|$$ is the spectral norm of $$A$$, in order that $$|A|^2=|A^T A|=c_1$$, the place $$c_1gecdotsge c_n$$ are the (nonnegative) eigenvalues of $$A^T A$$. We have
$$|det A|^2=det(A^T A)=c_1cdots c_nle c_1^n=|A|^{2n},$$
in order that (1) follows.

This reasoning too exhibits that the equality in (1) holds (iff $$c_1=cdots=c_n$$, that’s) iff $$A$$ is a actual a number of of an orthogonal matrix.

On the opposite hand, we will have $$det A=0$$ however $$|A|=1$$ (say).

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