# Volume of ellipsoid proportional to inverse of matrix

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?

• 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. – amd Nov 22 '18 at 1:24
• The larger $A$, the smaller the volume. So I do not see why you think the volume is proportional to det A. – LinAlg Nov 22 '18 at 2:40
• – sequence Nov 22 '18 at 3:08
• @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. – LinAlg Nov 22 '18 at 3:17
• @LinAlg Do you mean that this set contains less elements $x$ for bigger $a$? – sequence Nov 22 '18 at 3:26