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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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