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In a book on convex optimization it is said that if an ellipsoid is defined as

$$\mathcal{E} =\{v:\|Av-b\|_2\le 1\}$$

with $A\in S^n, A \succ 0$ ($A$ is a positive definite matrix and $b$ is a real vector), $b\in \mathbb{R}^n$, then the volume of the ellipsoid is proportional to $\det A^{-1}$. However, I know that the volume of an ellipsoid is proportional to $\det A$, not its inverse. What am I missing here?

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  • $\begingroup$ Seems to me that neither is true. The determinant is equal to the product of the eigenvalues, and the latter are the inverse squares of the ellipsoid’s semiaxis lengths. $\endgroup$ – amd Nov 22 '18 at 1:24
  • $\begingroup$ The larger $A$, the smaller the volume. So I do not see why you think the volume is proportional to det A. $\endgroup$ – LinAlg Nov 22 '18 at 2:40
  • $\begingroup$ @LinAlg math.stackexchange.com/questions/711410/… $\endgroup$ – sequence Nov 22 '18 at 3:08
  • $\begingroup$ @sequence The answer depends on the description of the ellipsoid. How do $a,b,c$ relate to $A$? I hope you agree that the set $\{x : |ax - 0| \leq 1\} \subset \mathbb{R}$ gets smaller as $a$ gets bigger. $\endgroup$ – LinAlg Nov 22 '18 at 3:17
  • $\begingroup$ @LinAlg Do you mean that this set contains less elements $x$ for bigger $a$? $\endgroup$ – sequence Nov 22 '18 at 3:26

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